Algebra

Quadratic Equation Solver

Enter the coefficients of a quadratic equation ax² + bx + c = 0. The solver reports the vertex and axis of symmetry, computes the discriminant Δ = b² − 4ac to classify the roots, and applies the quadratic formula x = (−b ± √Δ) / (2a), showing every substitution.

Quadratic Equation Solver

Solve ax² + bx + c = 0 with roots, vertex and discriminant.

Try:
Answerx₁ = 2, x₂ = 1
  1. Equation1x² − 3x + 2 = 0
  2. Vertex(h, k) = (−b/2a, c − b²/4a) = (1.5, -0.25)
  3. Axis of symmetryx = 1.5
  4. DiscriminantΔ = b² − 4ac = (-3)² − 4·1·2 = 1
  5. Formulax = (−b ± √Δ) / (2a)
  6. x₁(−(-3) + √1) / (2·1) = 2
  7. x₂(−(-3) − √1) / (2·1) = 1

Formula and method

x = (−b ± √(b² − 4ac)) / (2a), discriminant Δ = b² − 4ac

First compute the discriminant Δ = b² − 4ac to classify the roots. If Δ > 0 there are two distinct real roots; Δ = 0 gives one repeated root x = −b/(2a); Δ < 0 yields a pair of complex conjugates x = −b/(2a) ± i·√(−Δ)/(2a). The vertex (−b/2a, c − b²/4a) and axis of symmetry are reported alongside.

Worked examples

Key terms

Frequently asked questions

What is the discriminant?

It is Δ = b² − 4ac. Positive gives two real roots, zero gives one repeated root, and negative gives complex conjugate roots.

Does it find the vertex?

Yes. The vertex is at (−b/2a, c − b²/4a), and the axis of symmetry is the vertical line through its x-coordinate.

Can it handle complex roots?

Yes. When the discriminant is negative the roots are returned in the form x = p ± qi.