How to Solve a Quadratic Equation with the Formula

Get the equation into the right shape first

The formula only works on an equation written as ax² + bx + c = 0, with a zero on the right-hand side. Almost every mistake starts before the formula is even used, at this step.

If your equation is x² + 5x = 14, move everything to one side first: x² + 5x − 14 = 0. Now a = 1, b = 5, c = −14.

If it is 3x² = 12 − x, rearrange to 3x² + x − 12 = 0.

Until the right-hand side is zero, the numbers you read off will be the wrong ones.

Identify a, b and c — with their signs

Write down all three values before touching the formula, and keep the minus signs attached to them.

For x² − 2x − 15 = 0: a = 1, b = −2, c = −15.

This is where most errors happen. b is −2, not 2. Losing that minus sign changes the answer completely, and because the calculation still works you get a plausible-looking wrong result.

If a term is missing, its coefficient is 0. In x² − 9 = 0, b = 0. If x² has no number in front, a = 1.

The formula

x = (−b ± √(b² − 4ac)) ÷ 2a

The ± is what gives you two answers. You work the whole thing out once with a plus and once with a minus.

Notice that the entire top line is divided by 2a. A very common slip is to divide only the square root part, leaving the −b outside the fraction.

Substitute, using brackets

Put brackets around every value as you substitute, especially the negative ones. It looks fussy and it prevents a whole category of sign error.

For a = 1, b = −2, c = −15:

x = (−(−2) ± √((−2)² − 4(1)(−15))) ÷ 2(1)

Without brackets, −2² is ambiguous and many people read it as −4 rather than +4. With brackets, (−2)² is unmistakably 4.

The calculator on this page writes every substitution with brackets for exactly this reason.

Work out the discriminant first

Deal with the part under the square root before anything else: b² − 4ac.

(−2)² = 4. And 4(1)(−15) = −60. So 4 − (−60) = 4 + 60 = 64.

Two negatives making a plus is the step people most often get wrong here. Subtracting −60 is the same as adding 60.

Knowing the discriminant early also tells you what to expect: 64 is positive and a perfect square, so there will be two neat rational roots.

Finish the two branches

√64 = 8, and 2a = 2.

Plus branch: x = (2 + 8) ÷ 2 = 10 ÷ 2 = 5.

Minus branch: x = (2 − 8) ÷ 2 = −6 ÷ 2 = −3.

Note that −b is −(−2) = +2, which is why both branches start with a positive 2.

So the solutions are x = 5 and x = −3.

Always check by substituting back

This takes twenty seconds and catches almost everything.

Put x = 5 into the original: 5² − 2(5) − 15 = 25 − 10 − 15 = 0. Correct.

Put x = −3 in: (−3)² − 2(−3) − 15 = 9 + 6 − 15 = 0. Correct.

If a root does not give zero, you have a sign error somewhere — and it is almost always in b or in the discriminant.