Quadratic Equation Examples: 4 Methods With Full Worked Solutions
Quadratic equation examples appear in virtually every algebra course — from middle school through AP Calculus prep — and mastering them unlocks a whole tier of problem-solving ability. A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0, and every such equation has exactly two solutions (which may be equal, real, or complex). The challenge is knowing which method to reach for: factoring is fastest when the numbers cooperate, the quadratic formula always works, completing the square builds deep understanding, and graphing gives visual intuition. This guide works through real numerical quadratic equation examples for each method, from the simplest monic cases all the way to word problems and non-integer solutions, so you can recognize patterns quickly under exam conditions.
Contents
- 01What Is a Quadratic Equation? Core Concepts Before the Examples
- 02Quadratic Equation Examples Solved by Factoring
- 03Quadratic Equation Examples Using the Quadratic Formula
- 04Quadratic Equation Examples by Completing the Square
- 05Quadratic Equation Word Problem Examples
- 06Practice Problems: 6 Quadratic Equation Examples to Try Yourself
- 07Common Mistakes in Quadratic Equation Examples — and How to Fix Them
- 08When to Use Each Method: A Decision Guide
- 09Frequently Asked Questions About Quadratic Equation Examples
What Is a Quadratic Equation? Core Concepts Before the Examples
A quadratic equation is any polynomial equation of degree 2, meaning the highest power of the variable is 2. The standard form is ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The coefficient a is the leading coefficient, b is the linear coefficient, and c is the constant term. The word "quadratic" comes from the Latin quadratus, meaning square — it refers to the x² term that defines the degree. Every quadratic equation has exactly two solutions, counted with multiplicity: two distinct real roots when the discriminant b² − 4ac is positive, a repeated real root when it equals zero, and two complex conjugate roots when it is negative. The three most common forms you will encounter are standard form (ax² + bx + c = 0), vertex form (a(x − h)² + k = 0), and factored form (a(x − r₁)(x − r₂) = 0). Converting between forms is often the key to choosing the right solution method. For example, vertex form makes it trivial to identify the vertex of the parabola and solve for x by taking a square root, while factored form makes the roots immediately visible. Before jumping into quadratic equation examples, it also helps to know the discriminant shortcut: calculate Δ = b² − 4ac first. If Δ is a perfect square (0, 1, 4, 9, 16, 25 …), factoring will give a clean integer answer. If Δ is positive but not a perfect square, the quadratic formula will give an irrational answer. If Δ is negative, the roots are complex and the quadratic formula is the only route.
The discriminant Δ = b² − 4ac decides the method: Δ is a perfect square → try factoring first; Δ > 0 but not a perfect square → use the quadratic formula; Δ < 0 → roots are complex.
Quadratic Equation Examples Solved by Factoring
Factoring is the fastest method when the quadratic has integer roots. The core idea is to rewrite ax² + bx + c as a product of two binomials, then apply the zero-product property: if (x − r₁)(x − r₂) = 0, then x = r₁ or x = r₂. For monic quadratics where a = 1, the process reduces to finding two numbers whose product is c and whose sum is b. For non-monic quadratics where a ≠ 1, the AC method splits the middle term into two parts that can be grouped and factored separately. The worked examples below cover both cases. Recognizing when factoring is appropriate saves significant time on timed tests — if you spot that b² − 4ac is a perfect square within a few seconds of reading the problem, go straight to factoring.
1. Example 1 (a = 1, both roots positive) — x² − 7x + 12 = 0
Step 1: Write in standard form. The equation is already in standard form with a = 1, b = −7, c = 12. Step 2: Find two numbers with product = 12 and sum = −7. Factor pairs of 12: (−3, −4) → product = 12 ✓, sum = −7 ✓. Step 3: Write the factored form. (x − 3)(x − 4) = 0. Step 4: Apply zero-product property. x − 3 = 0 → x = 3; x − 4 = 0 → x = 4. Solutions: x = 3 or x = 4. Check x = 3: 9 − 21 + 12 = 0 ✓. Check x = 4: 16 − 28 + 12 = 0 ✓.
2. Example 2 (a = 1, roots of opposite sign) — x² + 2x − 15 = 0
Step 1: Standard form confirmed: a = 1, b = 2, c = −15. Step 2: Find two numbers with product = −15 and sum = 2. Factor pairs of −15: (−3, 5) → product = −15 ✓, sum = 2 ✓. Step 3: Factored form. (x − 3)(x + 5) = 0. Step 4: x = 3 or x = −5. Check x = 3: 9 + 6 − 15 = 0 ✓. Check x = −5: 25 − 10 − 15 = 0 ✓.
3. Example 3 (a = 1, one root is zero) — x² − 9x = 0
Step 1: The equation has no constant term (c = 0). Factor out x directly: x(x − 9) = 0. Step 2: Apply zero-product property. x = 0 or x − 9 = 0 → x = 9. Solutions: x = 0 or x = 9. Many students forget that x = 0 is a valid solution — always check the case where the variable itself equals zero when c = 0.
4. Example 4 (a ≠ 1, AC method) — 2x² + 7x + 3 = 0
Step 1: Identify a = 2, b = 7, c = 3. Compute AC = 2 × 3 = 6. Step 2: Find two numbers with product = 6 and sum = 7. That pair is (1, 6): 1 × 6 = 6 ✓, 1 + 6 = 7 ✓. Step 3: Split the middle term using these numbers. 2x² + 1x + 6x + 3 = 0. Step 4: Group and factor. x(2x + 1) + 3(2x + 1) = 0. Factor out the common binomial: (x + 3)(2x + 1) = 0. Step 5: Solutions. x = −3 or 2x + 1 = 0 → x = −½. Check x = −3: 2(9) + 7(−3) + 3 = 18 − 21 + 3 = 0 ✓. Check x = −½: 2(¼) + 7(−½) + 3 = ½ − 7/2 + 3 = 0 ✓.
When c = 0, always factor out x first. When a ≠ 1, use the AC method: multiply a × c, find a factor pair that sums to b, split the middle term, then group.
Quadratic Equation Examples Using the Quadratic Formula
The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) works for every quadratic equation without exception. It is derived by completing the square on the general form ax² + bx + c = 0 and is the method of last resort when factoring fails or when the roots are irrational. The formula produces exact answers — leaving the radical in simplified form — or decimal approximations when needed. The ± symbol means you compute two separate values: one using the plus sign and one using the minus sign. A common error is forgetting to divide the entire numerator (−b ± √Δ) by 2a, not just the radical part. The worked examples below include a case with two distinct irrational roots and a case with a repeated root.
1. Example 5 (Two distinct irrational roots) — x² − 4x + 1 = 0
Step 1: Identify a = 1, b = −4, c = 1. Step 2: Compute the discriminant. Δ = (−4)² − 4(1)(1) = 16 − 4 = 12. Since 12 is not a perfect square, use the quadratic formula. Step 3: Apply the formula. x = (−(−4) ± √12) / (2 × 1) = (4 ± √12) / 2. Step 4: Simplify √12 = √(4 × 3) = 2√3. So x = (4 ± 2√3) / 2 = 2 ± √3. Solutions: x = 2 + √3 ≈ 3.732 or x = 2 − √3 ≈ 0.268. Check x = 2 + √3: (2 + √3)² − 4(2 + √3) + 1 = (4 + 4√3 + 3) − 8 − 4√3 + 1 = 7 + 4√3 − 8 − 4√3 + 1 = 0 ✓.
2. Example 6 (Repeated root / perfect square trinomial) — 9x² − 12x + 4 = 0
Step 1: Identify a = 9, b = −12, c = 4. Step 2: Discriminant. Δ = (−12)² − 4(9)(4) = 144 − 144 = 0. A discriminant of zero means there is exactly one solution (a repeated root). Step 3: Apply the formula. x = (−(−12) ± √0) / (2 × 9) = 12 / 18 = 2/3. The equation has one solution: x = 2/3 (a repeated root). Note: you could also recognize 9x² − 12x + 4 = (3x − 2)² = 0, confirming x = 2/3 by factoring as a perfect square trinomial.
3. Example 7 (Non-integer coefficients) — 3x² + 5x − 2 = 0
Step 1: Identify a = 3, b = 5, c = −2. Step 2: Discriminant. Δ = 25 − 4(3)(−2) = 25 + 24 = 49. Since 49 = 7², factoring would also work here, but we demonstrate the formula. Step 3: Apply the formula. x = (−5 ± 7) / 6. Using +: x = (−5 + 7) / 6 = 2/6 = 1/3. Using −: x = (−5 − 7) / 6 = −12/6 = −2. Solutions: x = 1/3 or x = −2.
4. Example 8 (Complex roots) — x² + 2x + 5 = 0
Step 1: Identify a = 1, b = 2, c = 5. Step 2: Discriminant. Δ = 4 − 20 = −16. Since Δ < 0, the roots are complex (imaginary). Step 3: Apply the formula. x = (−2 ± √(−16)) / 2 = (−2 ± 4i) / 2 = −1 ± 2i. Solutions: x = −1 + 2i or x = −1 − 2i. These are complex conjugate pairs. The graph of y = x² + 2x + 5 never crosses the x-axis, which is consistent with having no real roots.
Quadratic formula memory trick: 'Negative b, plus or minus square root of b-squared minus 4ac, all over 2a.' Write the formula at the top of your paper before a test — it is worth every second.
Quadratic Equation Examples by Completing the Square
Completing the square is both a solution method and a conceptual tool — it converts any quadratic into vertex form a(x − h)² + k = 0, from which you can read off the vertex (h, k) of the parabola and solve by taking a square root. It is the method that proves the quadratic formula (the formula is derived by completing the square on the general form), and it is essential for converting equations of circles and parabolas in coordinate geometry. For a monic quadratic, the process involves adding and subtracting (b/2)² to create a perfect square on the left side. For a non-monic quadratic, divide through by a first. The worked examples below show both cases.
1. Example 9 (Monic quadratic) — x² + 6x + 5 = 0
Step 1: Move the constant to the right. x² + 6x = −5. Step 2: Compute (b/2)² = (6/2)² = 9. Add 9 to both sides. x² + 6x + 9 = −5 + 9 = 4. Step 3: Write the left side as a perfect square. (x + 3)² = 4. Step 4: Take the square root of both sides. x + 3 = ±√4 = ±2. Step 5: Solve. x = −3 + 2 = −1 or x = −3 − 2 = −5. Solutions: x = −1 or x = −5. Check x = −1: 1 − 6 + 5 = 0 ✓. Check x = −5: 25 − 30 + 5 = 0 ✓.
2. Example 10 (Non-monic) — 2x² − 8x + 6 = 0
Step 1: Divide every term by the leading coefficient 2. x² − 4x + 3 = 0. Step 2: Move the constant to the right. x² − 4x = −3. Step 3: Compute (b/2)² = (−4/2)² = 4. Add 4 to both sides. x² − 4x + 4 = −3 + 4 = 1. Step 4: Perfect square form. (x − 2)² = 1. Step 5: Take the square root. x − 2 = ±1. Step 6: Solve. x = 2 + 1 = 3 or x = 2 − 1 = 1. Solutions: x = 3 or x = 1.
3. Example 11 (Irrational result) — x² + 4x − 3 = 0
Step 1: Move the constant to the right. x² + 4x = 3. Step 2: (b/2)² = (4/2)² = 4. Add 4 to both sides. x² + 4x + 4 = 7. Step 3: (x + 2)² = 7. Step 4: Take the square root. x + 2 = ±√7. Step 5: Solve. x = −2 + √7 ≈ 0.646 or x = −2 − √7 ≈ −4.646. The irrational result here is exact — leave it as −2 ± √7 unless a decimal approximation is specifically requested.
The completing-the-square formula to memorize: add (b/2)² to both sides of x² + bx = −c to form (x + b/2)² = (b/2)² − c. Everything follows from there.
Quadratic Equation Word Problem Examples
Word problems involving quadratic equations typically fall into three categories: projectile motion (height of a thrown or falling object), area problems (a rectangle or frame with a given area), and number problems (two numbers with a given product and sum or difference). The key skill is translating the verbal description into a quadratic equation in standard form, then solving and interpreting only the physically meaningful solution. In projectile problems, negative time values are discarded. In area problems, negative dimensions are discarded. The worked examples below cover one problem from each category.
1. Example 12 (Projectile motion) — When does a ball hit the ground?
Problem: A ball is thrown upward from a height of 1.5 m with an initial velocity of 14 m/s. The height in metres after t seconds is h = −4.9t² + 14t + 1.5. When does the ball hit the ground? Step 1: Set h = 0. −4.9t² + 14t + 1.5 = 0. Step 2: Multiply both sides by −1 to get a positive leading coefficient. 4.9t² − 14t − 1.5 = 0. Step 3: Apply the quadratic formula. a = 4.9, b = −14, c = −1.5. Δ = (−14)² − 4(4.9)(−1.5) = 196 + 29.4 = 225.4. √225.4 ≈ 15.013. t = (14 ± 15.013) / (2 × 4.9) = (14 ± 15.013) / 9.8. Using +: t = 29.013 / 9.8 ≈ 2.96 s. Using −: t = −1.013 / 9.8 ≈ −0.10 s (rejected — time cannot be negative). Answer: The ball hits the ground after approximately 2.96 seconds.
2. Example 13 (Area problem) — Find the dimensions of a rectangle
Problem: The length of a rectangle is 3 cm more than twice its width. The area is 35 cm². Find the dimensions. Step 1: Let width = w cm, then length = (2w + 3) cm. Step 2: Write the area equation. w(2w + 3) = 35. Step 3: Expand and rearrange into standard form. 2w² + 3w − 35 = 0. Step 4: Apply the quadratic formula. a = 2, b = 3, c = −35. Δ = 9 + 280 = 289 = 17². x = (−3 ± 17) / 4. Using +: w = 14/4 = 3.5 cm. Using −: w = −20/4 = −5 (rejected — width cannot be negative). Answer: Width = 3.5 cm, Length = 2(3.5) + 3 = 10 cm. Check: 3.5 × 10 = 35 cm² ✓.
3. Example 14 (Number problem) — Two consecutive odd integers
Problem: The product of two consecutive odd integers is 143. Find both integers. Step 1: Let the first odd integer = n. The next consecutive odd integer = n + 2. Step 2: Write the product equation. n(n + 2) = 143. Step 3: Expand and rearrange. n² + 2n − 143 = 0. Step 4: Discriminant check. Δ = 4 + 572 = 576 = 24². Factoring or formula: n = (−2 ± 24) / 2. Using +: n = 22/2 = 11. Using −: n = −26/2 = −13. Both solutions are valid (odd integers): the pairs are 11 and 13, or −13 and −11. Check: 11 × 13 = 143 ✓ and (−13)(−11) = 143 ✓.
For every word problem: (1) define your variable, (2) write the equation, (3) solve, (4) reject any physically impossible solutions (negative length, negative time), (5) re-read the question to confirm you answered what was asked.
Practice Problems: 6 Quadratic Equation Examples to Try Yourself
The only way to get faster at solving quadratic equations is to work through problems without looking at the solution first. For each problem below, decide on your method (factoring, quadratic formula, or completing the square) before calculating. Answers and brief solutions are provided after each problem — but cover them and attempt the problem on your own first. The problems progress from straightforward monic factoring through to a word problem, mirroring the difficulty curve on most algebra tests.
1. Problem A — x² − 11x + 28 = 0 (Factor this)
Solution: Find two numbers with product = 28 and sum = −11. That pair is (−4, −7): (−4)(−7) = 28 ✓, (−4) + (−7) = −11 ✓. Factored form: (x − 4)(x − 7) = 0. Solutions: x = 4 or x = 7.
2. Problem B — x² + 10x + 25 = 0 (Perfect square trinomial)
Solution: Recognize 25 = 5² and 10 = 2 × 5. This is a perfect square trinomial: (x + 5)² = 0. Repeated root: x = −5. Discriminant check: Δ = 100 − 100 = 0 ✓.
3. Problem C — 4x² − 17x − 15 = 0 (Use the quadratic formula)
Solution: a = 4, b = −17, c = −15. Δ = 289 + 240 = 529 = 23². x = (17 ± 23) / 8. Using +: x = 40/8 = 5. Using −: x = −6/8 = −3/4. Solutions: x = 5 or x = −3/4.
4. Problem D — x² − 6x + 7 = 0 (Complete the square)
Solution: x² − 6x = −7. Add (6/2)² = 9 to both sides: (x − 3)² = 2. x = 3 ± √2. Exact solutions: x = 3 + √2 ≈ 4.414 or x = 3 − √2 ≈ 1.586.
5. Problem E — 3x² + x − 2 = 0 (AC method factoring)
Solution: AC = 3 × (−2) = −6. Find two numbers with product = −6 and sum = 1: that pair is (−2, 3). Split: 3x² − 2x + 3x − 2 = 0. Group: x(3x − 2) + 1(3x − 2) = 0. Factor: (x + 1)(3x − 2) = 0. Solutions: x = −1 or x = 2/3.
6. Problem F (Word problem) — Garden border
A square garden has side length x metres. A border of uniform width 2 m is added on all sides, making the total area 144 m². Find x. Setting up: the total side length is x + 4, so (x + 4)² = 144. Expand: x² + 8x + 16 = 144. Rearrange: x² + 8x − 128 = 0. Discriminant: 64 + 512 = 576 = 24². x = (−8 + 24) / 2 = 8 (take positive root). The garden is 8 m × 8 m. Check: (8 + 4)² = 144 ✓.
Before every quadratic problem, pause for five seconds: is c = 0 (factor out x), is Δ a perfect square (factor or perfect square trinomial), or do I need the formula? The three-second diagnostic saves minutes.
Common Mistakes in Quadratic Equation Examples — and How to Fix Them
Mistakes in quadratic equations usually fall into a small number of categories that repeat across students and exams. Knowing them in advance lets you build habits that avoid them automatically. The most frequent errors are sign errors when reading b and c from standard form, forgetting to divide the entire numerator by 2a in the quadratic formula, discarding valid negative solutions in pure math problems (negative solutions are only discarded in applied word problems where context forbids them), and failing to simplify the radical in the final answer. The table below lists the six most common mistakes alongside the correct approach.
1. Mistake 1 — Wrong sign on b or c
Error: From x² − 5x + 6 = 0, a student writes b = 5 instead of b = −5 and gets incorrect factor pairs. Fix: Always include the sign as part of the coefficient. b is whatever multiplies x, including its sign. In x² − 5x + 6, the term is −5x, so b = −5. A useful check: rewrite the equation on a new line before identifying a, b, c.
2. Mistake 2 — Dividing only the radical by 2a
Error: x = −b ± √Δ / (2a) written as if only √Δ is divided. The correct expression is (−b ± √Δ) / (2a) — the entire numerator is divided by 2a. Fix: Always use full parentheses: write the formula with a fraction bar under the entire numerator. A quick numerical check: for 2x² − 4x − 6 = 0, the roots should be x = 3 and x = −1. If your answer is different, check the denominator.
3. Mistake 3 — Stopping after one solution
Error: After applying the ± sign in the formula, a student only computes the + case and writes one answer. Fix: A quadratic equation always has two solutions (which may be equal). Always compute both the + and − cases explicitly, even if you suspect one will be rejected. Write them out separately: x₁ = (−b + √Δ)/(2a) and x₂ = (−b − √Δ)/(2a).
4. Mistake 4 — Forgetting to simplify the radical
Error: Leaving the answer as x = (4 ± √12) / 2 without simplifying √12 = 2√3, giving x = 2 ± √3. Fix: After computing the discriminant, always check whether it has a perfect square factor. Factor it: √12 = √(4 × 3) = 2√3. This matters because examiners expect simplified radical form, and un-simplified answers lose marks even when the setup is correct.
5. Mistake 5 — Discarding a valid negative solution
Error: In the problem 'find two numbers whose product is 12 and sum is −7', a student finds x = −3 and x = −4 but discards the negative solutions because 'numbers can't be negative'. Fix: Negative solutions are valid in pure algebra unless the problem specifies a real-world constraint (like length or time) that forbids them. Always re-read the question: if it asks for the numbers, negative integers are perfectly valid answers. Only discard negative values in applied problems where the context rules them out.
6. Mistake 6 — Wrong sign in factored form
Error: From the roots x = 3 and x = −5, a student writes the factored form as (x + 3)(x − 5) instead of (x − 3)(x + 5). Fix: If the root is x = r, the corresponding factor is (x − r). A positive root r gives the factor (x − r), which has a negative sign. A negative root r gives (x − r) = (x − (−|r|)) = (x + |r|), which has a positive sign. The sign in the factor is the opposite of the root.
Quick sanity check after solving: substitute both roots back into the original equation. If either check fails, there is a sign error or arithmetic slip somewhere — do not skip verification on exams.
When to Use Each Method: A Decision Guide
Choosing the right method for a quadratic equation example depends on the structure of the equation and what the problem is asking for. There is no single best method — each has contexts where it is fastest. The guide below is the decision logic that experienced algebra students use automatically after enough practice. Once you internalize this decision tree, you will rarely waste time on the wrong approach.
1. Decision 1 — Is c = 0?
If the constant term c = 0, factor out x immediately. For example, 5x² − 20x = 0 becomes x(5x − 20) = 0, giving x = 0 or x = 4. Do not use the quadratic formula here — it works, but factoring is far faster and the x = 0 root is obvious.
2. Decision 2 — Is it a special pattern?
Check for two special cases: (a) Difference of squares: if the equation is ax² − c = 0 with no middle term (b = 0), rewrite as (√a · x + √c)(√a · x − √c) = 0. Example: 4x² − 25 = 0 → (2x + 5)(2x − 5) = 0 → x = ±5/2. (b) Perfect square trinomial: if Δ = 0, the trinomial is a perfect square. Example: x² − 14x + 49 = (x − 7)².
3. Decision 3 — Is Δ a perfect square?
Compute Δ = b² − 4ac. If Δ is 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, or any other perfect square, factoring will give integer or simple fraction roots. Use the factor-pair method (for a = 1) or the AC method (for a ≠ 1). If Δ is positive but not a perfect square, the roots are irrational — use the quadratic formula.
4. Decision 4 — None of the above?
Use the quadratic formula. It always works. For decimals or word problems where you need a numerical approximation, calculate Δ first, then √Δ, then substitute. For problems requiring exact form (in coursework or proofs), simplify the radical as far as possible and leave the answer as (−b ± √Δ) / (2a) in simplified radical form.
Method selection order: (1) c = 0 → factor out x. (2) Special pattern → difference of squares or perfect square. (3) Δ is a perfect square → factor. (4) Everything else → quadratic formula.
Frequently Asked Questions About Quadratic Equation Examples
Students preparing for algebra tests consistently run into the same questions about quadratic equations. The answers below address the most common points of confusion, drawn from the types of errors that appear most frequently on homework and exams.
1. Q: Can a quadratic equation have just one solution?
Yes — when the discriminant Δ = b² − 4ac equals exactly zero, the two solutions coincide: x = −b/(2a). This is called a repeated root or double root. Geometrically, it means the parabola y = ax² + bx + c just touches the x-axis at one point (is tangent to it) without crossing. Example: x² − 6x + 9 = 0 has Δ = 36 − 36 = 0, giving the single solution x = 3.
2. Q: Why does my calculator give a different decimal than the exact answer?
When roots are irrational (like 2 + √3 or 3 − √7), any decimal approximation is rounded and will never exactly match a hand-calculated exact form. Always carry the exact form (simplified radical) in your working and only convert to a decimal at the end when the problem asks for it. On most standardized tests, exact form is required unless the problem says 'round to the nearest hundredth.'
3. Q: How do I know if a quadratic equation can be factored with integers?
Compute the discriminant Δ = b² − 4ac. If Δ is a perfect square (0, 1, 4, 9, 16, 25, 36 …), the equation can be factored over the integers (or rational numbers). If Δ is positive but not a perfect square, the roots are irrational — factoring with integers is impossible, and the quadratic formula gives exact irrational roots. If Δ < 0, the roots are complex numbers.
4. Q: What is the difference between a quadratic equation and a quadratic expression?
A quadratic expression (or quadratic polynomial) is just the algebraic expression ax² + bx + c without an equals sign — for example, x² + 5x + 6. A quadratic equation sets a quadratic expression equal to zero (or any constant): ax² + bx + c = 0. You solve equations (finding values of x); you factor or evaluate expressions. The distinction matters because 'solve x² + 5x + 6' is incomplete — you need an equals sign to solve. The correct form is 'solve x² + 5x + 6 = 0.'
5. Q: Do I need to learn all three methods or just the quadratic formula?
In practice, the quadratic formula is the one method that always works, so knowing it cold is non-negotiable. However, factoring is significantly faster for the majority of textbook problems (those with small integer coefficients) and demonstrates deeper algebraic understanding — most teachers and examiners reward it. Completing the square is tested explicitly in many courses because it reveals the vertex and is used to derive the quadratic formula. The practical answer: learn all three, default to factoring first on timed tests, and use the formula when factoring does not produce a clean answer quickly.
If you only have time to memorize one thing: x = (−b ± √(b² − 4ac)) / (2a). It solves every quadratic equation, every time.
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