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Quadratic Equation Solver

Solve ax² + bx + c = 0 with real or complex roots.

in-browser

How to use

  1. 1 Enter the coefficient a (the number in front of x²).
  2. 2 Enter b (the number in front of x) and c (the constant term).
  3. 3 Read the roots, discriminant, vertex and axis of symmetry below.
  4. 4 Copy the full solution if you need it for your notes.

About Quadratic Equation Solver

The Quadratic Equation Solver takes the three coefficients of a quadratic, ax² + bx + c = 0, and returns its roots using the quadratic formula.

Enter a, b and c — whole numbers, decimals or negatives all work — and the solver immediately shows you the answers, whether they are two distinct real numbers, a single repeated root, or a pair of complex conjugates.

It does more than spit out two numbers.

The tool reports the discriminant, b² − 4ac, which is the value that determines the nature of the roots: positive means two real solutions, zero means one repeated solution, and negative means the parabola never crosses the x-axis, so the roots are complex.

When the discriminant is negative the answers are shown in the familiar a ± bi form.

Alongside the roots you also get the vertex of the parabola and its axis of symmetry, x = −b ÷ 2a, which is useful for graphing or for optimisation problems.

Everything is calculated in your browser with plain arithmetic — no equation, value or session is ever sent to a server — so it is fast, works offline once loaded, and is well suited to homework, exam revision and engineering checks where you simply need a trustworthy answer.

FAQ

What does the discriminant tell me?

The discriminant b² − 4ac decides the roots: greater than zero gives two real roots, exactly zero gives one repeated root, and less than zero gives a complex conjugate pair.

Why can a not be zero?

If a is zero the x² term disappears and the expression is linear, not quadratic. Use a linear approach instead, since the quadratic formula divides by a.

Can it handle complex roots?

Yes. When the discriminant is negative the solver returns the roots in a ± bi form rather than failing.