Quadratic Equation Solver

Solve ax² + bx + c = 0 with full step-by-step working.

Enter coefficients for ax² + bx + c = 0

a (x²)

b (x)

c (constant)

1x² − 5x + 6 = 0

Discriminant (b²−4ac)1

Root typetwo real roots

Vertex (h, k)(2.5, -0.25)

Axis of symmetryx = 2.5

x = 3

x = 2

Step-by-step

1. Discriminant = b² − 4ac = (-5)² − 4·(1)·(6) = 1

2. √discriminant = 1

3. x = (−b ± √discriminant) / 2a = (−(-5) ± 1) / 2

Quadratic Formula

Any equation of the form ax² + bx + c = 0 (where a ≠ 0) is solved with the quadratic formula: x = (−b ± √(b²−4ac)) / 2a. The discriminant (b²−4ac) determines the nature of roots: positive → two real roots, zero → one repeated root, negative → two complex roots.

Built and maintained by Meet Shah · Last updated 19 July 2026

What this tool is used for

Solving a quadratic and seeing the discriminant that determines the root type.
Finding a vertex and an axis of symmetry for a graph.
Reading the discriminant to know the root type before solving at all.
Getting complex roots when the discriminant is negative.
Confirming a factorisation you found by inspection.

Frequently Asked Questions

What does the discriminant tell me before I see the roots?: Everything about their nature. b² − 4ac is the quantity under the square root: positive gives two distinct real roots, exactly zero gives one repeated root where the parabola just touches the x-axis, and negative gives a conjugate pair of complex roots and a curve that never crosses the axis at all.
Why does entering a = 0 produce no result?: Because it is not a quadratic. With a = 0 the equation collapses to bx + c = 0, a straight line with the single root −c/b, and the formula itself divides by 2a — so the case is refused rather than returning Infinity. The tool requires a ≠ 0, which is part of the definition of a quadratic.
How is the vertex calculated?: The x-coordinate is −b / 2a, midway between the two roots because a parabola is symmetric about its axis. The y-coordinate uses c − b²/(4a) directly rather than substituting x back in, which avoids compounding the rounding from the first step. For x² − 5x + 6 the vertex is (2.5, −0.25).
What does a complex root actually mean here?: That the parabola sits entirely above or below the x-axis. The roots still exist — they always come as a conjugate pair p ± qi, with the real part p equal to the vertex's x-coordinate — but no real input makes the expression zero. In circuit and control problems that pair is what encodes an oscillation.
Is the quadratic formula always the most accurate method?: Not always. When b² is far larger than 4ac, one of the two additions cancels almost exactly and loses significant digits — a known instability that Vieta's x₁x₂ = c/a sidesteps. For everyday coefficients it never surfaces, but it is why numerical libraries avoid the schoolbook formula.

Common errors and gotchas

Getting the sign of b wrong in the formula, which is the most common slip.
Assuming real roots when the discriminant is negative.
Dividing by 2a and forgetting a can be negative, which flips the parabola.
Losing precision when b is large relative to the discriminant, where the standard formula is numerically poor.
Treating a repeated root as two distinct ones.