A quadratic equation ax² + bx + c = 0 is solved by the quadratic formula: x = (−b ± √(b²−4ac)) / (2a). The quantity under the root, the discriminant D = b²−4ac, classifies the solutions before you even finish: D > 0 gives two real roots, D = 0 a single repeated root, and D < 0 a pair of complex-conjugate roots. The vertex of the parabola sits at x = −b/(2a), halfway between the roots. Together these tell you everything about the equation's shape and solutions.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the quadratic formula and Vieta's relations.
The quadratic equations
The ± in the formula produces the two roots, symmetric about the vertical line x = −b/(2a). The discriminant is the deciding quantity: a positive value has a real square root (two crossings of the x-axis), zero means the root sits exactly on the axis, and a negative value forces an imaginary part, giving conjugate complex roots. Vieta's relations — sum = −b/a, product = c/a — provide a quick independent check on any answer.
Worked example — a factorable quadratic
Scenario: solve x² − 5x + 6 = 0 (a = 1, b = −5, c = 6).
With D = 1 the roots are real and distinct: 3 and 2. The vertex sits at x = 2.5 — exactly midway between them — with y = −0.25. The checks hold: the roots sum to 5 (= −b/a) and multiply to 6 (= c/a). Other equations show the other cases: x² − 4x + 4 has D = 0 and a single root of 2; x² + x + 1 has D = −3 and complex roots −0.5 ± 0.866i; and x² − 2x − 3 has D = 16 with roots 3 and −1.
Frequently Asked Questions
x = (−b ± √(b²−4ac)) / (2a). x²−5x+6 → x = (5 ± 1)/2 = 3, 2.
D = b²−4ac. >0 two real, =0 one repeated, <0 two complex roots.
No real roots — two complex conjugates p ± qi. x²+x+1 → −0.5 ± 0.866i.
x = −b/(2a), y = c − b²/(4a). x²−5x+6 → (2.5, −0.25).
Vieta: sum of roots = −b/a, product = c/a. 3+2=5, 3×2=6. ✓