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x² − 5x + 6 = 0 → x = 3 and x = 2, discriminant 1, vertex at (2.5, −0.25). Two roots, one root, or complex — all three cases shown.
The quadratic formula never fails — factoring does. x = (−b ± √(b²−4ac)) / 2a solves every quadratic, and the discriminant (b²−4ac) tells you what kind of answer to expect before you compute anything. This calculator returns both roots, the discriminant with its meaning, the parabola's vertex, and complex roots in a ± bi form when the curve never touches the x-axis.
Algebra students use it nightly: x² − 5x + 6 → x = 3 and x = 2, discriminant 1, vertex (2.5, −0.25) — all three discriminant cases handled, complex roots included.
Everything about the roots before you solve: b²−4ac > 0 means two distinct real roots (the parabola crosses the x-axis twice), = 0 means one repeated real root (it touches the axis exactly at the vertex), < 0 means two complex conjugate roots (it never crosses). The discriminant also controls whether factoring is possible over the integers: roots are rational only when the discriminant is a perfect square. x²−5x+6 has discriminant 1 — a perfect square — so it factors as (x−2)(x−3). x²−4x+1 has discriminant 12 — not a perfect square — so the roots 2±√3 are irrational and no integer factoring exists. Check the discriminant first and you know whether to factor, complete the square, or go straight to the formula.
Factor when it is quick — the formula when it is not. Factoring is faster and builds intuition when the discriminant is a perfect square and coefficients are small: x²−5x+6 becomes (x−2)(x−3) in seconds. But the formula works on everything, always: 2x²+4x−4 has roots 0.732 and −2.732 that no amount of staring would factor. The professional habit: compute the discriminant first. Perfect square → factor or formula, your choice. Not a perfect square → formula directly. Negative → complex roots, done. The two sanity checks that catch 90% of errors: the roots must sum to −b/a and multiply to c/a. For x²−5x+6: roots 3 and 2 sum to 5 = −(−5)/1 and multiply to 6 = 6/1. Ten seconds, errors eliminated.
The vertex is the parabola's turning point — the minimum when a > 0 (cup opens up) or maximum when a < 0 (opens down) — sitting at x = −b/2a, y = c − b²/4a. It is the answer to every optimization word problem: a ball's peak height, maximum revenue, minimum cost. A projectile following h = −16t² + 64t + 6 peaks at t = −64/(2·−16) = 2 seconds, reaching h = 70 feet — no calculus needed. The vertex also anchors the graph: the parabola is symmetric about the vertical line through it, so knowing the vertex plus one point gives you the whole curve. For x²−5x+6 the vertex is (2.5, −0.25) — the curve bottoms out a quarter-unit below the axis, exactly between its roots at 2 and 3, as symmetry demands.