How to Solve Any Quadratic Equation: The Formula, the Discriminant, and the Parabola

A quadratic equation is any equation you can write as ax² + bx + c = 0, where a is not zero. That last clause matters: if a is zero, the x² term vanishes and you are left with a straight line, bx + c = 0, which is a different problem entirely. Everything below assumes a real, nonzero a.

Most students meet quadratics around age 14 or 15 and then keep running into them for years: projectile motion in physics, optimization in economics, the path of anything that gets thrown or dropped. The good news is that one method solves every single quadratic with no guessing. The interesting news is that the method also tells you, before you finish, exactly what kind of answer you are about to get.

The quadratic formula does everything

The formula is:

x = (−b ± √(b² − 4ac)) / (2a)

Plug in a, b, and c, and out come the roots. The ± means there are usually two of them — one for the plus branch, one for the minus branch. People sometimes carry only one sign and quietly lose half the answer, so write both every time.

The formula never fails. Factoring fails the moment your numbers stop being friendly; completing the square works but takes more steps. The formula just grinds through any coefficients you hand it. If you want to double-check the arithmetic on messy decimals, a scientific calculator handles the square root and division without rounding errors creeping in.

The discriminant decides everything before you solve

Look at the part under the square root: b² − 4ac. This is the discriminant, written Δ (delta). Its sign alone tells you how many real roots exist, and you can read it off in one second without finishing the calculation.

Δ > 0: the square root is a real positive number, the ± splits it into two values, and you get two distinct real roots.
Δ = 0: the square root is zero, so both branches land on the same value. You get one repeated real root (sometimes called a double root).
Δ < 0: the square root of a negative number is not real. It pulls out an imaginary unit i, and you get a conjugate pair of complex roots, p + qi and p − qi.

That third case is why some textbook problems answer "no real solutions". The equation still has solutions — they just live in the complex plane. Because a, b, and c are real, complex roots always come in conjugate pairs: same real part, opposite imaginary parts. It is a property of every real-coefficient polynomial, not a quirk of quadratics.

There is a bonus. If Δ is a perfect square (1, 4, 9, 16, …), the roots are rational and the equation factors cleanly over the integers. That single check tells you whether factoring is even worth attempting.

A worked example: solving x² − 5x + 6 = 0

Here a = 1, b = −5, c = 6.

First the discriminant:

Δ = b² − 4ac = (−5)² − 4(1)(6) = 25 − 24 = 1

Δ is positive, so two distinct real roots. It is also a perfect square (1), so the roots are rational and the equation will factor. Now the formula:

x = (5 ± √1) / 2 = (5 ± 1) / 2

The plus branch gives (5 + 1) / 2 = 3. The minus branch gives (5 − 1) / 2 = 2. So the roots are x = 2 and x = 3.

You can sanity-check with Vieta's formulas, which say the sum of the roots equals −b/a and the product equals c/a. Sum: 2 + 3 = 5 = −(−5)/1. Product: 2 × 3 = 6 = 6/1. Both match, so the answer is solid. And because Δ was a perfect square, the factored form x² − 5x + 6 = (x − 2)(x − 3) drops right out.

Factoring vs the formula: when to use which

Factoring is faster — when it works. If you can spot two numbers that multiply to a·c and add to b, you are done in seconds. For x² − 5x + 6 the numbers −2 and −3 jump out, so factoring beats grinding the formula.

The trap is spending thirty seconds hunting for factors that do not exist. This is exactly where the discriminant earns its keep: compute Δ first. Perfect square means factoring will work, so go for it. Anything else (a non-square positive number, or a negative number) means the roots are irrational or complex, and no integer factoring will land them. Reach for the formula and stop guessing.

When I tutor this, the habit I push hardest is "discriminant first, method second." I watched a student burn most of a timed test trying to factor 2x² + 3x − 4, an equation whose Δ is 41 — not a perfect square, so it never could factor over the integers. One discriminant check up front would have saved her the whole detour and sent her straight to the formula. That one reflex changes quadratics from a fishing expedition into a procedure.

The parabola and its vertex

Every quadratic ax² + bx + c is also a curve: a parabola. The equation ax² + bx + c = 0 asks where that curve crosses the x-axis, and those crossing points are precisely the real roots. Now the three discriminant cases turn into pictures:

Δ > 0: the parabola cuts the x-axis at two points (two real roots).
Δ = 0: it just touches the axis at one point — the vertex sits exactly on it.
Δ < 0: it never reaches the axis, floating entirely above or below it. No real roots, which is why the curve and the x-axis never meet.

The turning point of the parabola is the vertex, located at x = −b/2a. Its y-value is the maximum or minimum of the whole function: a minimum if a > 0 (the parabola opens upward) and a maximum if a < 0 (it opens downward). For x² − 5x + 6, the vertex sits at x = 5/2 = 2.5, with y = −0.25 — the lowest point of the curve, tucked just below the axis between the roots at 2 and 3.

This is why parabola problems show up everywhere. A ball thrown into the air follows h(t) = −5t² + 20t + 1; the vertex (t = 2, h = 21) gives the maximum height and when it happens, no calculus required. The vertex formula does the work.

Watching all of this at once — discriminant, both roots, vertex, and the drawn curve — makes the structure click in a way that a single number cannot. The quadratic equation solver lays out the formula, completing-the-square, and Vieta's check side by side, then sketches the parabola with the vertex and intercepts marked, so you see why each answer comes out the way it does.

Quick recap

To solve any ax² + bx + c = 0: compute Δ = b² − 4ac first. Its sign tells you whether you get two real roots, one repeated root, or a complex pair. A perfect-square Δ means factor; anything else means reach for the formula. The vertex at x = −b/2a gives you the turning point and tells you which way the parabola opens. Master those three ideas and quadratics stop being a guessing game.

Made by Toolora · Updated 2026-06-13