A worked example: x² − 5x + 6 = 0
Take x² − 5x + 6 = 0, so a = 1, b = -5, and c = 6. The Quadratic Formula Calculator first works out the discriminant, b² − 4ac, which comes to (-5)² − 4(1)(6) = 25 − 24 = 1. Because that discriminant is positive, the equation has two distinct real roots. The calculator returns root1 = 3 and root2 = 2, and indeed 3² − 5(3) + 6 = 0 and 2² − 5(2) + 6 = 0.
The formula: applying x = (−b ± √(b² − 4ac)) ÷ 2a
The quadratic formula itself is x = (−b ± √(b² − 4ac)) ÷ 2a. For 2x² − 7x + 3 = 0 (a = 2, b = -7, c = 3), the discriminant is (-7)² − 4(2)(3) = 49 − 24 = 25, and √25 = 5. Substituting gives x = (7 ± 5) ÷ 4, which the calculator reports as root1 = 3 and root2 = 0.5 — the plus sign gives (7 + 5) ÷ 4 = 3, and the minus sign gives (7 − 5) ÷ 4 = 0.5.
Two more examples: a repeated root and a complex pair
When the discriminant is exactly zero, the ± term vanishes and both roots collapse to the same value. For 4x² − 12x + 9 = 0 (a = 4, b = -12, c = 9), the discriminant is (-12)² − 4(4)(9) = 144 − 144 = 0, and the calculator reports "One repeated real root" with root1 = root2 = 1.5.
When the discriminant is negative, there is no real square root, so the calculator switches to complex output instead of failing. For x² + 2x + 5 = 0 (a = 1, b = 2, c = 5), the discriminant is 2² − 4(1)(5) = 4 − 20 = -16. The real part of each root is −b ÷ 2a = -2 ÷ 2 = -1, and the imaginary part is √16 ÷ 2 = 2, so the calculator returns "Two complex conjugate roots": root1 = "-1 + 2i" and root2 = "-1 - 2i".
Why the discriminant's sign matters
The discriminant alone tells you what kind of answer to expect before you even finish the arithmetic — positive means the parabola crosses the x-axis twice, zero means it just touches the axis at a single point, and negative means it never touches the axis at all, so the roots only exist in the complex plane. Checking that sign first is a quick sanity check on whichever root values a calculator or a hand computation produces.