Quadratic Formula

The quadratic formula solves a quadratic equation in standard form:

It gives the value or values of $x$ that make the quadratic equal to zero, where $a$, $b$, and $c$ are the coefficients in $ax^2 + bx + c = 0$.

When this method is the right choice

Use the quadratic formula for any equation of the form $ax^2 + bx + c = 0$ with $a \ne 0$. It is most useful when:

1. A quadratic does not factor cleanly. If a quadratic factors quickly, factoring is often faster, but when it does not, the formula is the reliable method that still works.
2. You want a method that always works for standard-form quadratics.
3. You want to know how many real solutions to expect from the discriminant.
4. You are comparing methods such as factoring, completing the square, and graphing.

The part under the square root,

is called the discriminant. It predicts the kind of answer before you finish the arithmetic:

1. If $b^2 - 4ac > 0$, there are two distinct real solutions.
2. If $b^2 - 4ac = 0$, there is one repeated real solution.
3. If $b^2 - 4ac < 0$, there are no real solutions. In that case, the solutions are complex.

A quadratic can have up to two $x$-values where its graph crosses the $x$-axis. The formula is the general result of completing the square, so it gives those intercepts directly without guessing factors.

The procedure, step by step

1. Put the equation in standard form. Rewrite it so all terms are on one side and the quadratic equals zero.
2. Identify the coefficients. Read off $a$, $b$, and $c$, paying close attention to negative signs.
3. Substitute into the formula. Simplify the discriminant before taking the square root.
4. Evaluate both cases. The $\pm$ means you work out both the plus case and the minus case, since they can give different solutions.

A full example through every step

Solve $2x^2 + 3x - 2 = 0$.

Step 1 and 2 — standard form and coefficients. The equation is already in standard form, so

Step 3 — substitute.

Work inside the square root first:

So the formula becomes

Step 4 — both branches.

So the solutions are

Check one root by substitution. When $x = \frac{1}{2}$,

That confirms the value works.

Where each step tends to break, and how to check

At standard form: not rewriting the equation as $ax^2 + bx + c = 0$ first. If the right side is not zero, the coefficients are not ready for the formula. Self-check: is one side exactly $0$?

At identifying coefficients: losing the sign of $b$ or $c$. If $b = -7$, then $-b = 7$, not $-7$. Self-check: copy each sign deliberately.

At substituting: forgetting that the denominator is the whole $2a$. The entire numerator $-b \pm \sqrt{b^2 - 4ac}$ sits over $2a$. Also watch for arithmetic errors inside the discriminant, since small sign mistakes there change the entire answer.

At evaluating: calculating only one case. The $\pm$ requires both versions. Self-check: substitute one root back into the original equation, as in the worked example.

Build the skill on a new equation

Solve $x^2 - 6x + 5 = 0$ with the same steps: put it in standard form, identify $a$, $b$, and $c$, compute the discriminant, and evaluate both branches. For a useful comparison, factor it afterward and check that both methods give the same roots.