Quadratic equation solver

x = (−b ± √(b² − 4ac)) ÷ 2a. Discriminant D = b² − 4ac. Vertex at x = −b ÷ 2a, y = c − b² ÷ 4a. Write the equation in the form ax² + bx + c = 0 and enter the three coefficients. The discriminant decides the kind of answer: positive gives two different real roots, zero gives one repeated root, and negative gives a pair of complex roots of the form p ± qi. The roots are computed in a numerically stable way, so equations with very different coefficient sizes keep their accuracy.

With a (coefficient of x²) 2, b (coefficient of x) -4, c (constant term) -6, the solutions is x = 3 or x = -1. Discriminant (b² − 4ac): 64; Root 1 (real part): 3; Root 2 (real part): -1; Imaginary part: 0; Vertex x (axis of symmetry): 1; Vertex y (turning point): -8.

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D = b² − 4ac. If D is positive the parabola crosses the x-axis twice (two real roots); if D is zero it touches the axis once (a repeated root); if D is negative it never reaches the axis and the roots are complex.

Here a = 2, b = −4 and c = −6, so D = 16 + 48 = 64 and √D = 8. The roots are (4 ± 8) ÷ 4, which gives x = 3 or x = −1.

Without the x² term the equation becomes bx + c = 0, a straight line with at most one solution, x = −c ÷ b. The quadratic formula would divide by zero.