The Quadratic Formula: How to Solve ax² + bx + c = 0
x = (−b ± √(b²−4ac)) / 2a solves any quadratic equation written in the form ax² + bx + c = 0. This guide works through the formula step by step, explains the discriminant, and covers when to use the formula instead of factoring.

The quadratic formula is
x = (−b ± √(b²−4ac)) / 2a. Plug in the coefficients a, b, and c from an equation in standard form and it returns every solution. It never fails, which is why it is worth memorizing.
Want the answer without the algebra? The quadratic formula calculator takes a, b, and c and shows the roots plus each step. The rest of this page explains where the formula comes from and how to read its output.
What the formula says
A quadratic written in standard form has three coefficients: a multiplies x², b multiplies x, and c is the constant term. The formula returns the solutions for that form, on the single condition that a is not zero. If a were zero there would be no x² term and the equation would not be quadratic.
The ± symbol is what gives a quadratic its two answers. It tells you to compute the expression twice, once adding the square root and once subtracting it. The two results are usually the two solutions.
Working an example with two real roots
Take a clean equation already in standard form and read off the coefficients first, then move through the formula one piece at a time.
Start with the part under the root. Compute the discriminant, then take its square root.
Now apply the ± over the denominator and split. The denominator is twice the leading coefficient, and each sign gives one root.
Both answers check out when substituted back. Putting each root into the original equation returns zero, which confirms the work.
What the discriminant tells you
The piece under the square root is called the discriminant, written D. Its sign tells you how many real solutions exist before you finish any arithmetic, because you cannot take the real square root of a negative number and the square root of zero is just zero.
D > 0 → two distinct real roots
D = 0 → one repeated real root
D < 0 → two complex (conjugate) rootsThe earlier example had a positive discriminant, which is why it split into two separate answers. When the discriminant is zero instead, the ± adds and subtracts nothing, leaving a single value.
Working an example with complex roots
When the discriminant comes out negative, the roots are complex. Read off the coefficients and compute the discriminant the same way as before.
A negative discriminant pulls in the imaginary unit. Writing the square root of the negative discriminant in terms of i, where i is the square root of negative one, gives a conjugate pair.
The two solutions share the same real part and have opposite imaginary parts, which is the conjugate pair the table predicts whenever the discriminant is negative. The form is always p ± qi.
When to use the formula instead of factoring
Factoring is faster when it works. If you can spot two numbers that multiply to ac and add to b, you can split the middle term and read off the roots in seconds. The earlier example factors directly, no formula required.
two numbers multiply to 6, add to −5 → −2 and −3
(x − 2)(x − 3) = 0Reach for the formula when factoring stalls. A few situations make factoring impractical, and the formula handles all of them without guessing:
- The roots are not whole numbers or simple fractions.
- The discriminant is negative, so there is nothing to factor over the reals.
- The leading coefficient is not one and the cross-multiplication gets messy.
- You simply want a method that always works.
Completing the square is a third option, and it is in fact the derivation the formula comes from. Once you have the formula memorized, it is rarely worth doing by hand.
The formula is the one method that solves every quadratic, so when the numbers are messy or the roots turn complex, skip the guesswork and plug your coefficients into the quadratic formula calculator for the discriminant, both roots, and the worked steps.
Common questions
It is the expression under the square root, D = b² − 4ac. Its sign tells you whether the equation has two real roots, one repeated root, or two complex roots, without computing the roots themselves.
When the discriminant is negative you take the square root of a negative number, so the roots use the imaginary unit i = √(−1). Geometrically, the parabola never crosses the x-axis, so no real value of x makes it zero. The solutions still exist as the conjugate pair p ± qi.
Then the squared term disappears and the equation is linear, solved by x = −c / b. The quadratic formula does not apply, because its denominator would then be zero. Always confirm that a is not zero before reaching for the formula.


