The Quadratic Formula, Step by Step

The quadratic formula is the one method that solves every quadratic equation, even the ones that cannot be factored neatly. It can look intimidating written out as a single long expression, but it is really just four arithmetic steps repeated the same way every time. This walkthrough covers the formula itself, three worked examples of increasing difficulty, and how to check your own answer, with every result verified by substituting it back into the original equation.

The formula and what each part means

For any quadratic equation in the standard form ax² + bx + c = 0 (where a is not zero), the quadratic formula states:

x = (−b ± √(b² − 4ac)) / (2a)

The Common Core high school algebra standards group this under "Reasoning with Equations and Inequalities," specifically the expectation that students can "solve quadratic equations in one variable," including by using the quadratic formula. The three letters a, b, and c are not abstract; they are simply the numbers sitting in front of x², x, and the constant term in your specific equation, read off in that order.

The expression under the square root, b² − 4ac, is called the discriminant, and it tells you in advance what kind of answer to expect:

• Positive discriminant: two different real solutions.
• Zero discriminant: exactly one real solution (the two "±" branches give the same number).
• Negative discriminant: no real solutions (the equation's graph does not cross the x-axis).

Example 1: a simple two-solution case

Solve x² − 5x + 6 = 0.

Here a = 1, b = −5, c = 6. Substitute directly into the formula:

x = (5 ± √((−5)² − 4·1·6)) / (2·1) = (5 ± √(25 − 24)) / 2 = (5 ± √1) / 2 = (5 ± 1) / 2

This gives two answers: x = (5+1)/2 = 3, and x = (5−1)/2 = 2.

Checking the answer: substituting x = 3 back into the original equation gives 3² − 5(3) + 6 = 9 − 15 + 6 = 0. Substituting x = 2 gives 2² − 5(2) + 6 = 4 − 10 + 6 = 0. Both check out exactly.

Example 2: an equation with a leading coefficient other than 1

Solve 2x² + 3x − 2 = 0.

Here a = 2, b = 3, c = −2. Substitute:

x = (−3 ± √(3² − 4·2·(−2))) / (2·2) = (−3 ± √(9 + 16)) / 4 = (−3 ± √25) / 4 = (−3 ± 5) / 4

This gives x = (−3+5)/4 = 0.5, and x = (−3−5)/4 = −2.

Checking the answer: 2(0.5)² + 3(0.5) − 2 = 2(0.25) + 1.5 − 2 = 0.5 + 1.5 − 2 = 0. For x = −2: 2(−2)² + 3(−2) − 2 = 2(4) − 6 − 2 = 8 − 6 − 2 = 0. Both confirmed.

A common mistake in this example is forgetting that subtracting a negative c flips the sign inside the formula; notice that −4ac became −4(2)(−2), which is +16, not −16. Watch the sign of c carefully whenever it is negative.

Example 3: a repeated (double) root

Solve x² + 4x + 4 = 0.

Here a = 1, b = 4, c = 4. Substitute:

x = (−4 ± √(4² − 4·1·4)) / (2·1) = (−4 ± √(16 − 16)) / 2 = (−4 ± 0) / 2 = −2

Because the discriminant is exactly zero, both branches of the "±" give the same number, so there is only one solution: x = −2. This matches the discriminant rule above (zero discriminant means exactly one real solution).

Checking the answer: (−2)² + 4(−2) + 4 = 4 − 8 + 4 = 0. Confirmed.

What a negative discriminant tells you

Not every quadratic equation has a real-number solution. Take x² + 2x + 5 = 0: here a = 1, b = 2, c = 5, so the discriminant is 2² − 4(1)(5) = 4 − 20 = −16, which is negative. Since you cannot take the square root of a negative number within the real numbers, this equation has no real solutions. Recognizing this before you start simplifying saves time; if your discriminant comes out negative, there is no real x value to report, and you do not need to keep working the problem as though there is.

A reliable process for any quadratic equation

1. Rewrite the equation in standard form (ax² + bx + c = 0) if it is not already, moving every term to one side.
2. Identify a, b, and c directly from the equation, paying close attention to signs.
3. Calculate the discriminant (b² − 4ac) first, before plugging into the rest of the formula, so you immediately know whether to expect two solutions, one, or none.
4. Substitute into the full formula and simplify carefully, handling the square root and the division by 2a as separate steps rather than trying to do everything at once.
5. Check your answer by substituting it back into the original equation. If the left and right sides do not come out equal, go back and recheck your arithmetic, particularly your handling of negative signs.

Key takeaways

• The quadratic formula, x = (−b ± √(b² − 4ac)) / (2a), solves any equation in the form ax² + bx + c = 0, including ones that cannot be factored.
• Calculate the discriminant (b² − 4ac) first: positive means two solutions, zero means one solution, negative means no real solutions.
• Negative values of b or c are a common source of sign errors; double-check the arithmetic inside the square root and the numerator separately.
• Always verify a solution by substituting it back into the original equation; this step catches nearly every arithmetic mistake before it becomes a wrong final answer.
• A discriminant of exactly zero produces a repeated (double) root, meaning both "±" branches of the formula simplify to the same number.