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Quadratic Equation Solver

Solve quadratic equations — free and step by step with the quadratic formula. Enter a, b and c and the calculator shows the discriminant and both solutions.

Quick answer
How do you solve a quadratic equation?
Put the equation in the form ax² + bx + c = 0 and substitute into the quadratic formula: x = (−b ± √(b² − 4ac)) / (2a). The discriminant D = b² − 4ac decides the number of solutions: D > 0 gives two, D = 0 gives one, D < 0 gives no real solution. Example x² − 5x + 6 = 0: D = 25 − 24 = 1, so x = (5 ± 1)/2, giving x₁ = 3 and x₂ = 2.
The tool

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Comma or dot as decimal separator, negative values allowed.
Step-by-step
Press Calculate to see every step.
HowTo

Solve a quadratic equation — 4 steps

Worked through “x² − 5x + 6 = 0”
  1. 1
    Step 1 of 4

    Put it in standard form

    Write the equation as ax² + bx + c = 0. Here a = 1, b = −5, c = 6.

  2. 2
    Step 2 of 4

    Compute the discriminant

    D = b² − 4ac = (−5)² − 4 · 1 · 6 = 25 − 24 = 1. D > 0 means two solutions.

  3. 3
    Step 3 of 4

    Substitute into the quadratic formula

    x = (−b ± √D) / (2a) = (5 ± √1) / 2 = (5 ± 1)/2.

  4. 4
    Step 4 of 4

    Work out both solutions

    x₁ = (5 + 1)/2 = 3 and x₂ = (5 − 1)/2 = 2. Check: 3 + 2 = 5 = −b/a, 3 · 2 = 6 = c/a — correct.

Examples

Quadratic equations — worked examples

Using the quadratic formula and the discriminant
x² − 5x + 6 = 0
D = 25 − 24 = 1
x = (5 ± 1)/2
x₁ = 3, x₂ = 2
x² − 4 = 0
D = 0 + 16 = 16
x = (0 ± 4)/2
x₁ = 2, x₂ = −2
x² + 6x + 9 = 0
D = 36 − 36 = 0
x = −6/2
x = −3 (repeated)
2x² − 7x + 3 = 0
D = 49 − 24 = 25
x = (7 ± 5)/4
x₁ = 3, x₂ = 0.5
x² + x + 1 = 0
D = 1 − 4 = −3
D < 0
no real solution
3x² − 12x = 0
D = 144 − 0 = 144
x = (12 ± 12)/6
x₁ = 4, x₂ = 0
Theory

The quadratic formula and the discriminant

A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. Its solutions come from the quadratic formula: x = (−b ± √(b² − 4ac)) / (2a). The expression under the square root, D = b² − 4ac, is called the discriminant and controls the solution set: when D > 0 there are two distinct real solutions, when D = 0 there is one repeated solution (the parabola touches the x-axis), and when D < 0 there is no real solution (the parabola never crosses the x-axis). Many equations can also be solved faster by factoring or with Vieta’s formulas — which double as a handy check: x₁ + x₂ = −b/a and x₁ · x₂ = c/a. Quadratic equations appear from grade 9 onward and underpin parabolas, projectile motion and optimization problems.

Pitfalls

Common mistakes when solving

Forgetting the sign of b

The formula uses −b. When b = −5 that becomes +5. Use brackets: (−5)² = 25, not −25.

Not in standard form

Move everything to one side (ax² + bx + c = 0) before reading off a, b, c.

Misreading the discriminant

D < 0 means no real solution — not “x = 0”. D = 0 gives exactly one solution.

Forgetting to divide by a

The denominator is 2a, not 2. When a = 2 you divide by 4.
FAQ

Frequently asked questions about quadratic equations

Glossary

Glossary — key terms explained simply

Quadratic equation
An equation of the form ax² + bx + c = 0 with a ≠ 0.
Quadratic formula
The solution formula x = (−b ± √(b² − 4ac)) / (2a).
Discriminant
The term D = b² − 4ac that determines the number of solutions.
Repeated root
A solution counted twice; occurs when D = 0.
Standard form
The form ax² + bx + c = 0 with everything on one side.
Vieta’s formulas
x₁ + x₂ = −b/a and x₁ · x₂ = c/a — handy for checking.