Quadratic Formula Calculator
What are the roots of x² − 5x + 6 = 0? Enter a, b, and c to solve any quadratic with the formula — real or complex roots, the discriminant, the vertex, and the substituted steps.
x² − 5x + 6 = 0
Two real roots
x = 3 or x = 2
Discriminant
1
b² − 4ac
Vertex
(2.5, −0.25)
(−b/2a, c − b²/4a)
Axis of symmetry
x = 2.5
vertical line
Steps
x = (−b ± √(b² − 4ac)) / 2a b² − 4ac = (−5)² − 4 × 1 × 6 = 1 x = (5 ± √1) / 2 = (5 ± 1) / 2 x = 3 or x = 2
The ± in the formula is the parabola's symmetry: both roots sit the same distance from the axis x = −b/2a, so the vertex is always exactly halfway between them. The discriminant b² − 4ac decides everything — positive means two crossings of the x-axis, zero means the vertex touches it, negative means the parabola misses it entirely.
The Discriminant Decides Everything
For x² − 5x + 6 = 0 the discriminant is b² − 4ac = 25 − 24 = 1, so the roots are (5 ± 1) ÷ 2 — that is, x = 3 and x = 2 — with the vertex halfway between at (2.5, −0.25). Make the discriminant negative and the parabola never touches the x-axis: x² + 4x + 5 = 0 has D = −4, giving the complex pair −2 ± i. And the formula scales with any leading coefficient — 2x² + 4x − 6 = 0 has D = 64, giving x = 1 and x = −3.
The formula & the three cases
x = (−b ± √(b² − 4ac)) / 2a D > 0 two real roots x² − 5x + 6: D = 1 → x = 3, x = 2 D = 0 one repeated root vertex sits on the x-axis D < 0 complex pair p ± qi x² + 4x + 5: D = −4 → x = −2 ± i Vertex: (−b/2a, c − b²/4a) Axis of symmetry: x = −b/2a
Frequently Asked Questions
What does a negative discriminant mean?
No real solutions — the parabola never crosses the x-axis. The roots become a complex conjugate pair: x² + 4x + 5 = 0 has discriminant 16 − 20 = −4, giving x = −2 ± i. The real part is −b/2a and the imaginary part is √|D|/2a, so the pair always mirrors across the axis of symmetry.
Can I solve a quadratic without the quadratic formula?
Often — x² − 5x + 6 factors as (x − 2)(x − 3), giving the roots 2 and 3 directly, and completing the square also works. But factoring only cooperates when the roots are tidy rational numbers; the formula is the method that never fails, including for irrational and complex roots.
How do I find the vertex of a parabola from a, b, and c?
The vertex sits on the axis of symmetry at x = −b/2a, with y = c − b²/4a. For x² − 5x + 6 that's x = 5/2 = 2.5 and y = 6 − 25/4 = −0.25, so the vertex is (2.5, −0.25) — exactly halfway between the roots 2 and 3, and the parabola's minimum since a > 0.
What happens if a = 0 in ax² + bx + c = 0?
It stops being quadratic — the x² term vanishes and you're left with the linear equation bx + c = 0, which has the single root x = −c/b (when b ≠ 0). The quadratic formula itself breaks at a = 0 because it divides by 2a, which is why the calculator switches to the linear solution instead.