Quadratic Equation Solver

The Formula

x = (−b ± √(b²−4ac)) ÷ 2a. The ± gives two solutions (roots). When the discriminant (b²−4ac) is positive: two real roots. When zero: one repeated root. When negative: two complex roots.

Example

Worked example: solve x² − 5x + 6 = 0. Identify a=1, b=−5, c=6. Calculate discriminant: b² − 4ac = 25 − 24 = 1. Since discriminant > 0, there are two distinct real roots. Apply the quadratic formula: x = (−b ± √discriminant) / (2a) = (5 ± 1) / 2. Root 1: x = (5+1)/2 = 3. Root 2: x = (5−1)/2 = 2. Verify by substituting: 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓ and 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓. Both values satisfy the original equation.

Applications

Quadratic equations model projectile motion, area problems, profit maximisation, and countless physics problems. The roots represent where the parabola y = ax² + bx + c crosses the x-axis.

Real-World Quadratic Applications

Quadratic equations appear wherever relationships involve squared variables. Projectile motion: height h = ut - 1/2 gt squared. Revenue maximisation: revenue is a quadratic function of price that peaks at the optimal price-quantity combination. Area problems: finding dimensions of a rectangle with given area and perimeter constraints. The discriminant (b squared minus 4ac) determines the number of solutions: positive gives two real solutions, zero gives one repeated solution, and negative gives