Mega Quadratic Equations Solver

Take (x − 2)(x + 3) = 5x − 1. Expand the left: x² + 3x − 2x − 6, which collects to x² + x − 6. The right is already 5x − 1. Nothing is in standard form yet. Subtract the right from the left: x² + x − 6 − 5x + 1 = x² − 4x − 5. So the equation is x² − 4x − 5 = 0, and only now can the formula be applied. The discriminant is (−4)² − 4(1)(−5) = 16 + 20 = 36. Thirty-six is a square number, so the roots are rational and the quadratic factors: x² − 4x − 5 = (x − 5)(x + 1), giving x = 5 and x = −1. Check both in the original equation rather than in the rearranged one. At x = 5: the left is (3)(8) = 24 and the right is 25 − 1 = 24. At x = −1: the left is (−3)(2) = −6 and the right is −5 − 1 = −6. Both hold. Now one that does not factor: x² − 3x − 5 = 0. The discriminant is 9 + 20 = 29, which is not a square number, so the roots are (3 ± √29)/2 — about 4.1926 and −1.1926. No amount of searching will find integer factors here, because there are none. That is the case for knowing the test: the discriminant tells you whether factoring is even possible before you start looking. And one where the square disappears: (x + 1)² = x² + 5. Expanding the left gives x² + 2x + 1. Subtracting the right: x² + 2x + 1 − x² − 5 = 2x − 4. The squared terms cancelled exactly, so the equation is 2x − 4 = 0 and x = 2. It was never a quadratic. Checking in the original: (3)² = 9, and 4 + 5 = 9. One last thing worth knowing about the formula itself. For x² + 1000000000x + 25 = 0 the small root is about −2.5 × 10⁻⁸, but computing it as (−b + √(b² − 4ac))/2a in ordinary floating point subtracts two nearly equal numbers and returns something closer to −6 × 10⁻⁸ — or, at larger coefficients, exactly zero. Across a family built to test this, the textbook arrangement failed to satisfy its own equation 90% of the time. Working in exact fractions avoids the problem rather than mitigating it.