How to Use the Quadratic Formula: A Beginner’s Walk-Through

The quadratic formula is the one tool that solves any quadratic equation, even the ones that won’t factor. Memorize it, learn to use it cleanly, and you can handle quadratics on the SAT, ACT, GED, and beyond.

The formula

For any quadratic in the form \(ax^2 + bx + c = 0\):

\[x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]

The \(\pm\) gives you two answers (the two roots of the parabola).

The 4-step routine

1. Get the equation in standard form (\(ax^2 + bx + c = 0\)).
2. Identify \(a\), \(b\), and \(c\).
3. Plug into the formula, carefully with signs.
4. Simplify the square root, then split into two answers.

Worked example

Solve \(2x^2 – 5x – 3 = 0\).

1. Standard form: ✓.
2. \(a = 2\), \(b = -5\), \(c = -3\).
3. Plug in: \(x = \dfrac{-(-5) \pm \sqrt{(-5)^2 – 4(2)(-3)}}{2(2)} = \dfrac{5 \pm \sqrt{25 + 24}}{4} = \dfrac{5 \pm \sqrt{49}}{4} = \dfrac{5 \pm 7}{4}\).
4. Two answers: \(x = \tfrac{12}{4} = 3\) or \(x = \tfrac{-2}{4} = -\tfrac{1}{2}\).

The discriminant, peek before you solve

The expression under the square root, \(b^2 – 4ac\), is called the discriminant. It tells you what kind of roots you’ll get:

• Positive → two distinct real solutions.
• Zero → one repeated real solution.
• Negative → two complex solutions (no real roots).

Use it to know in advance whether your answer will be clean.

When to use the formula vs. Factoring

Common mistakes

• Forgetting the negative on \(-b\).
• Squaring \(b\) as \(-b^2\) instead of \(b^2\). (\((-5)^2 = 25\), not $-25$.)
• Forgetting the \(\pm\), and missing half your answer.

FAQ

What is the quadratic formula?

The quadratic formula is x = (−b ± √(b² − 4ac)) / (2a). It gives the solutions of any quadratic equation written in standard form, ax² + bx + c = 0, where a is not 0. The ± means you usually get two answers, one using plus and one using minus. When those answers are real numbers, they are the x-intercepts of the parabola y = ax² + bx + c.

When do you use the quadratic formula?

Use it whenever a quadratic will not factor easily, or when you are not sure it factors at all, because it always works. For x² − 7x + 12 = 0, factoring into (x − 3)(x − 4) is faster. But for x² − 2x − 1 = 0, no whole numbers work, and the formula gives x = 1 ± √2. A good habit is to look for an easy factor first, then switch to the formula.

How do you use the quadratic formula to solve an equation?

Write the equation in standard form, identify a, b and c, substitute them into the formula, and simplify. For 3x² + 5x + 2 = 0, a = 3, b = 5 and c = 2. The discriminant is 5² − 4(3)(2) = 25 − 24 = 1, so x = (−5 ± 1)/6. That gives x = −4/6 = −2/3 or x = −6/6 = −1. Plug one answer back in to check.

What does the discriminant tell you in the quadratic formula?

The discriminant is b² − 4ac, the part under the square root. If it is positive, there are two real solutions. If it is zero, there is one repeated real solution. If it is negative, there are no real solutions, only complex ones, and the parabola never touches the x-axis. For x² + 2x + 5 = 0, the discriminant is 4 − 20 = −16, so there are no real solutions.

Is the quadratic formula the same as factoring?

No. They are two different ways to solve the same kind of equation. Factoring rewrites ax² + bx + c as a product, like (x − 3)(x − 4), and is quick only when the numbers are friendly. The quadratic formula skips that search and always gives the solutions, even irrational or complex ones. When factoring is possible, both methods give the same answers.

What is the quadratic formula derived from?

It comes from completing the square on the general equation ax² + bx + c = 0. You divide every term by a, move c/a to the other side, add (b/2a)² to both sides, and rewrite the left side as (x + b/2a)². Taking the square root of both sides and solving for x produces x = (−b ± √(b² − 4ac)) / (2a). That is why the formula works for every quadratic.

Worked example with a negative discriminant

Solve \(x^2 + 2x + 5 = 0\).

• \(a = 1\), \(b = 2\), \(c = 5\).
• Discriminant: \(4 – 20 = -16\).
• \(\sqrt{-16} = 4i\).
• \(x = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i\).

Two complex roots: \(-1 + 2i\) and $-1 – 2i$. The parabola never crosses the x-axis.

Worked example with a repeated root

Solve \(x^2 – 6x + 9 = 0\).

• \(a = 1\), \(b = -6\), \(c = 9\).
• Discriminant: \(36 – 36 = 0\).
• \(x = \dfrac{6 \pm 0}{2} = 3\).

One repeated root: \(x = 3\). The parabola just touches the x-axis at a single point. Always double-check by noticing the perfect-square trinomial: \(x^2 – 6x + 9 = (x – 3)^2\).

Where quadratics appear in the real world

• Projectile motion. A ball thrown straight up follows the equation \(h = -16t^2 + v_0 t + h_0\). Solving for when it lands (\(h = 0\)) is a quadratic-formula problem.
• Profit maximization. Revenue × quantity sold often produces a quadratic; the vertex tells you the price that maximizes profit.
• Geometry. Diagonal lengths, areas, and Pythagorean problems often collapse into quadratics.
• Physics. Energy, motion, and oscillation formulas frequently involve \(x^2\).

In every case, the quadratic formula is your universal tool.

Test-day cheat sheet

• Always check first if the quadratic factors easily.
• If factoring fails, write down \(a\), \(b\), \(c\) carefully (watch the signs).
• Compute the discriminant first. It tells you what kind of answer to expect.
• Don’t forget the \(\pm\). Two answers, almost always.

Vieta’s formulas, a hidden shortcut

For a quadratic \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\):

• Sum of roots: \(r_1 + r_2 = -\dfrac{b}{a}\)
• Product of roots: \(r_1 \cdot r_2 = \dfrac{c}{a}\)

This lets you check your answers fast. For \(x^2 – 5x + 6 = 0\), the roots should sum to 5 and multiply to 6. The roots are 2 and 3. ✓

Vieta’s also lets you build a quadratic given its roots: if the roots are 4 and $-3$, then \(b = -(4 + (-3))(1) = -1\) and \(c = 4 \cdot (-3) = -12\), so the quadratic is \(x^2 – x – 12 = 0\).

Comparing all 4 methods

Which method should you use? Here’s a decision tree:

1. Can you factor it in 10 seconds? Use factoring.
2. Is \(b\) even and \(a = 1\)? Try completing the square. It’ll be clean.
3. Is the equation already \((\text{something})^2 = \text{number}\)? Take square roots directly.
4. Otherwise: use the quadratic formula. It always works.

The quadratic formula is the universal hammer. The other methods are faster when they apply.

Practice set

Solve each:

1. \(x^2 – 7x + 12 = 0\)
2. \(2x^2 + 5x – 3 = 0\)
3. \(x^2 + 4x + 5 = 0\)
4. \(3x^2 – 6x + 2 = 0\)
5. \(x^2 – 9 = 0\)
6. \(4x^2 = 12x\)

Answers: 1) \(x = 3, 4\). 2) \(x = \tfrac{1}{2}, -3\). 3) \(x = -2 \pm i\). 4) \(x = 1 \pm \tfrac{\sqrt{3}}{3}\). 5) \(x = \pm 3\). 6) \(x = 0, 3\).

Final note

The quadratic formula isn’t scary. It’s just a very compressed recipe. Write it down at the start of every test. Plug in the numbers carefully. Double-check signs. You’ll be right far more often than the students who try to memorize tricks.

Extra study tips that move the needle

Most students don’t fail because the math is too hard. They fail because their practice habits are inefficient. Here are the habits that separate the students who improve fast from those who stall.

Practice with a timer. Untimed practice teaches you to eventually get the right answer; timed practice teaches you to get it in test conditions. Set a stopwatch every time you sit down. Aim for 90 seconds per question on most standardized tests.

Keep an error log. A simple spreadsheet with three columns, Problem, My answer, Correct answer, Why I missed it, is the single most powerful study tool ever invented. Review your error log weekly. The same mistakes show up again and again until you name them.

Mix topics every session. Doing 20 problems on the same topic feels productive, but spaced and interleaved practice, mixing topics, builds retrieval skills, which is what the test actually measures. Spend 70% of your time on mixed sets and only 30% on isolated drills.

Sleep on it. Memory consolidation happens during sleep. A 30-minute session the night before a quiz, followed by 7+ hours of sleep, beats a 3-hour cram session that ends at midnight. This is settled cognitive science.

Teach the topic out loud. If you can’t explain it, you don’t fully know it. Either record yourself, write a one-paragraph “how I’d teach this” explanation, or grab a friend to listen. Teaching exposes the gaps your problem sets hid.

When to ask for help

Spinning your wheels for more than 15 minutes on a single problem is a signal, not of failure, but of a missing piece of background. Stop, mark the problem, and either ask a teacher, post in our community, or watch a video on the relevant subtopic. Resuming after gaining the missing piece is much more efficient than guessing your way forward.

A quick self-assessment

Before you close this tab, answer these three questions honestly:

1. What’s the one topic in this article you understood best?
2. What’s the one topic that still feels fuzzy?
3. What concrete next step (a worksheet, a practice test, a video) will you take in the next 48 hours?

Writing those answers down, even just in a notes app, has been shown to roughly double the chance you actually follow through. Treat the next 48 hours as a small, doable experiment, not a marathon. Your future test-day self will thank you.

Drill it with our Algebra 1 worksheets or the complete Algebra Bundle.