How to Solve Quadratic Equations

A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. The exponent 2 on the leading term is what makes it quadratic — it produces a parabola when graphed, and up to two solutions (roots) when solved algebraically.

The quadratic formula

The universal method for solving any quadratic equation is the quadratic formula:

x = (−b ± √(b² − 4ac)) / (2a)

The ± means you get two candidate answers: one with addition and one with subtraction. Whether both are real numbers depends on the discriminant.

The discriminant (Δ = b² − 4ac)

Before computing both roots, evaluate Δ = b² − 4ac. Its sign tells you everything about the nature of the solutions:

Discriminant	Root type	What you get
Δ > 0	Two distinct real roots	Two different x-values where the parabola crosses the x-axis
Δ = 0	One repeated real root	The parabola touches the x-axis at exactly one point
Δ < 0	Two complex conjugate roots	No real x-intercepts; roots involve the imaginary unit i

Step-by-step example (two real roots)

Solve x² − 5x + 6 = 0 (a = 1, b = −5, c = 6).

Step 1 — Compute the discriminant:

Δ = (−5)² − 4(1)(6) = 25 − 24 = 1

Δ > 0, so two distinct real roots.

Step 2 — Apply the formula:

x = (−(−5) ± √1) / (2 × 1) = (5 ± 1) / 2

Step 3 — Evaluate both roots:

x₁ = (5 + 1) / 2 = 3
x₂ = (5 − 1) / 2 = 2

You can verify by substituting back: 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓

Step-by-step example (complex roots)

Solve x² + x + 1 = 0 (a = 1, b = 1, c = 1).

Step 1 — Discriminant:

Δ = 1² − 4(1)(1) = 1 − 4 = −3

Δ < 0, so two complex conjugate roots.

Step 2 — Formula with imaginary unit:

x = (−1 ± √(−3)) / 2 = (−1 ± i√3) / 2

These are valid mathematical answers — they just don't appear as x-intercepts on the real number line.

The degenerate case (a = 0)

When a = 0, the equation becomes bx + c = 0 — a linear equation with one solution: x = −c/b. The quadratic formula is undefined at a = 0 because it would divide by zero.

Other solving methods

The quadratic formula always works, but two shorter methods apply in special cases:

Factoring — if you can spot two numbers that multiply to ac and add to b. Example: x² + 5x + 6 = (x+2)(x+3) = 0 → roots are −2 and −3.
Completing the square — rewrite as (x + p)² = q, then take the square root of both sides. This is how the quadratic formula is derived.

For decimal coefficients, negative values, or discriminants that don't simplify neatly, the formula is always the safest path.

Quick reference

Situation	Check	Result
Integer roots expected	b² − 4ac is a perfect square	Factor instead
Coefficient a is large	Divide whole equation by a first	Simpler arithmetic
Roots are complex	Δ < 0	Express with i
a = 0	Linear, not quadratic	Use x = −c/b

Use the Quadratic Equation Solver to compute roots, the discriminant, and a full step-by-step working for any set of real coefficients — including cases with complex roots.