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Solve ax² + bx + c = 0 with the quadratic formula. Enter the three coefficients to get the roots — real or complex — along with the discriminant, vertex and every step.
x = (−b ± √(b² − 4ac)) ÷ 2a Δ = b² − 4ac
The discriminant Δ decides the type of roots: Δ > 0 gives two real roots, Δ = 0 one repeated root, and Δ < 0 two complex conjugate roots.
Solve 2x² + 3x − 5 = 0.
The graph of y = ax² + bx + c is a parabola. Its real roots are where it crosses the x-axis, and its vertex at x = −b ÷ 2a is the highest or lowest point. When a > 0 the parabola opens upwards; when a < 0 it opens downwards.
Then the equation is linear (bx + c = 0), not quadratic, and has the single solution x = −c ÷ b when b ≠ 0.
The parabola never crosses the x-axis, so there are no real roots. The two solutions are complex numbers of the form p ± qi.
Yes, when the roots are rational. x² − 3x + 2 = (x − 1)(x − 2), giving x = 1 and x = 2. The quadratic formula works for every case, including irrational and complex roots.
Results are computed in double-precision floating point and shown to six decimal places, using a numerically stable version of the formula.