Solving Quadratic Equations: Three Methods with Worked Examples

A quadratic equation is any equation of the form ax² + bx + c = 0. This guide explains three ways to solve them, factoring, completing the square, and the quadratic formula, with step-by-step worked examples and guidance on when to use each method.

What Is a Quadratic Equation?

A quadratic equation is a 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 constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.

Quadratic equations appear throughout mathematics, physics, engineering, and economics. They model projectile motion (the path of a ball), the area of a rectangle given its perimeter, profit maximisation in business, and the behaviour of electrical circuits.

The Discriminant: How Many Solutions?

Before choosing a solving method, the discriminant tells you what kind of solutions to expect:

Discriminant D = b² − 4ac D > 0 → two distinct real roots D = 0 → one repeated real root (the parabola touches the x-axis) D < 0 → two complex conjugate roots (no real solutions)

For example, x² − 5x + 6: D = 25 − 24 = 1 > 0 → two real roots. For x² + 1 = 0: D = 0 − 4 = −4 < 0 → complex roots only.

Method 1: Factoring

Factoring is the fastest method when it works. The goal is to rewrite ax² + bx + c as a product of two linear factors (px + q)(rx + s). Setting each factor to zero gives the roots.

When to use: When the coefficients are small integers and the discriminant is a perfect square. Not reliable for equations with irrational or complex roots.

Factoring Example, x² − 5x + 6 = 0

Find two numbers that multiply to c (6) and add to b (−5). The numbers are −2 and −3 because: (−2) × (−3) = 6 and (−2) + (−3) = −5 Factored form: (x − 2)(x − 3) = 0 Set each factor to zero: x − 2 = 0 → x = 2 x − 3 = 0 → x = 3 Solutions: x = 2 and x = 3

Factoring Example, 2x² + 5x − 3 = 0 (a ≠ 1)

Multiply a × c: 2 × (−3) = −6 Find two numbers that multiply to −6 and add to b (5): The numbers are 6 and −1 because: 6 × (−1) = −6 and 6 + (−1) = 5 Split the middle term and factor by grouping: 2x² + 6x − x − 3 = 0 2x(x + 3) − 1(x + 3) = 0 (2x − 1)(x + 3) = 0 Solutions: x = 1/2 and x = −3

Method 2: Completing the Square

Completing the square transforms the equation into the form (x + p)² = q, which can be solved by taking a square root. This method always works and is the foundation for deriving the quadratic formula.

When to use: When factoring fails but you want to understand the vertex form of the parabola, or when deriving the vertex (h, k) from y = ax² + bx + c.

Completing the Square, x² + 6x − 7 = 0

Step 1: Move the constant to the right side. x² + 6x = 7 Step 2: Add (b/2)² to both sides. b = 6, so (b/2)² = (3)² = 9 x² + 6x + 9 = 7 + 9 = 16 Step 3: Write the left side as a perfect square. (x + 3)² = 16 Step 4: Take the square root of both sides. x + 3 = ±4 Step 5: Solve for x. x = −3 + 4 = 1 x = −3 − 4 = −7 Solutions: x = 1 and x = −7

Completing the Square with a ≠ 1, 2x² − 8x + 3 = 0

Step 1: Divide through by a = 2. x² − 4x + 3/2 = 0 Step 2: Move the constant. x² − 4x = −3/2 Step 3: Add (b/2)² = (−2)² = 4 to both sides. x² − 4x + 4 = −3/2 + 4 = 5/2 Step 4: Perfect square form. (x − 2)² = 5/2 Step 5: Square root and solve. x − 2 = ±√(5/2) = ±√10 / 2 x = 2 ± √10 / 2 Solutions: x ≈ 2 + 1.581 = 3.581 and x ≈ 2 − 1.581 = 0.419

Method 3: The Quadratic Formula

The quadratic formula works for every quadratic equation without exception. It is derived by completing the square on the general form ax² + bx + c = 0:

x = (−b ± √(b² − 4ac)) / (2a)

When to use: Always, especially when factoring is not obvious or when coefficients are large, fractional, or irrational.

Quadratic Formula Example 1, 3x² − 4x − 5 = 0

a = 3, b = −4, c = −5 Discriminant: D = (−4)² − 4(3)(−5) = 16 + 60 = 76 x = (4 ± √76) / 6 = (4 ± 2√19) / 6 = (2 ± √19) / 3 x₁ = (2 + 4.359) / 3 ≈ 2.120 x₂ = (2 − 4.359) / 3 ≈ −0.786

Quadratic Formula Example 2, Complex Roots: x² + 4x + 5 = 0

a = 1, b = 4, c = 5 Discriminant: D = 16 − 20 = −4 (negative → complex roots) x = (−4 ± √(−4)) / 2 = (−4 ± 2i) / 2 = −2 ± i Solutions: x = −2 + i and x = −2 − i

Choosing the Right Method

Situation	Best method
Small integer coefficients, D is a perfect square	Factoring, fastest
Need vertex form or graph insight	Completing the square
Any equation, especially irrational or complex roots	Quadratic formula, always works
D < 0 (complex roots needed)	Quadratic formula
Approximate numerical answer acceptable	Quadratic formula or a calculator

Vieta's Formulas: A Quick Sanity Check

For a quadratic x² + bx + c = 0 with roots x₁ and x₂, Vieta's formulas let you verify your answers without substituting back into the equation:

Sum of roots: x₁ + x₂ = −b/a Product of roots: x₁ × x₂ = c/a Example: x² − 5x + 6 = 0, roots x = 2 and x = 3 Sum: 2 + 3 = 5 = −(−5)/1 ✓ Product: 2 × 3 = 6 = 6/1 ✓

Use our free Quadratic Equation Solver to solve any ax² + bx + c = 0 instantly, it shows the discriminant, both roots (including complex), step-by-step working, and a graph of the parabola.