How to Solve a Quadratic Equation? (+FREE Worksheet!)

Solving a quadratic equation means finding the values of \(\color{blue}{x}\) that make \(\color{blue}{\text{ ax }^{2} + \text{ bx } + c = 0}\) true. There are three main methods — factoring, the square root method, and the quadratic formula — and knowing all three gives you the flexibility to handle any quadratic you encounter in Algebra 1 and beyond.

What Is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree 2 in the standard form \(\color{blue}{\text{ ax }^{2} + \text{ bx } + c = 0}\), where \(\color{blue}{a \ne 0}\). The solutions (also called roots or zeros) are the x-values where the parabola \(\color{blue}{y = \text{ ax }^{2} + \text{ bx } + c}\) crosses the x-axis.

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Methods for Solving a Quadratic Equation

Method 1: Factoring

If the quadratic can be factored, write it as a product of two binomials set to zero, then apply the Zero Product Property: if \(\color{blue}{\text{ AB } = 0}\), then \(\color{blue}{A = 0}\) or \(\color{blue}{B = 0}\).

• \(\color{blue}{x^{2} + 5x + 6 = 0}\)  →  \(\color{blue}{(x + 2)(x + 3) = 0}\)  →  \(\color{blue}{x = -2}\) or \(\color{blue}{x = -3}\)

Method 2: Square Root Method

Use this when there is no \(\color{blue}{\text{ bx }}\) term (i.e., \(\color{blue}{b = 0}\)). Isolate the squared term and take the square root of both sides.

• \(\color{blue}{x^{2} – 25 = 0}\)  →  \(\color{blue}{x^{2} = 25}\)  →  \(\color{blue}{x = \pm 5}\)
• \(\color{blue}{2x^{2} = 50}\)  →  \(\color{blue}{x^{2} = 25}\)  →  \(\color{blue}{x = \pm 5}\)

Method 3: The Quadratic Formula

Works for any quadratic equation. Substitute \(\color{blue}{a}\), \(\color{blue}{b}\), and \(\color{blue}{c}\) into:

\(\color{blue}{x = (-b \pm \sqrt{(b^{2} – 4\text{ ac })}) \div (2a)}\)

• \(\color{blue}{2x^{2} + 5x – 3 = 0}\): \(\color{blue}{a=2, b=5, c=-3}\).
  Disc = \(\color{blue}{25 + 24 = 49}\). \(\color{blue}{x = \frac{(-5 \pm 7)}{4}}\).
  \(\color{blue}{x = \frac{2}{4} = \frac{1}{2}}\) or \(\color{blue}{x = -\frac{12}{4} = -3}\).

Step-by-Step Summary

1. Set the equation equal to zero (standard form).
2. Try factoring first — it is quickest when it works.
3. If no \(\color{blue}{\text{ bx }}\) term, use the square root method.
4. Otherwise, use the quadratic formula with \(\color{blue}{a}\), \(\color{blue}{b}\), \(\color{blue}{c}\) from standard form.
5. Check each solution by substituting it back into the original equation.

Watch: Introduction to Solving Quadratic Equations

Khan Academy introduces the quadratic equation and its solutions:

Solving a Quadratic Equation – Worked Examples

Example 1 (Factoring): Solve \(\color{blue}{x^{2} + 5x + 6 = 0}\).

Factor: \(\color{blue}{(x + 2)(x + 3) = 0}\).
Set each factor to zero: \(\color{blue}{x + 2 = 0}\)  →  \(\color{blue}{x = -2}\)     or     \(\color{blue}{x + 3 = 0}\)  →  \(\color{blue}{x = -3}\).
Solutions: \(\color{blue}{x = -2}\) and \(\color{blue}{x = -3}\).

Example 2 (Square Root): Solve \(\color{blue}{x^{2} – 49 = 0}\).

\(\color{blue}{x^{2} = 49}\)  →  \(\color{blue}{x = \pm 7}\).
Solutions: \(\color{blue}{x = 7}\) and \(\color{blue}{x = -7}\).

Example 3 (Quadratic Formula): Solve \(\color{blue}{x^{2} – 7x + 12 = 0}\).

\(\color{blue}{a=1, b=-7, c=12}\). Discriminant: \(\color{blue}{49 – 48 = 1}\).
\(\color{blue}{x = \frac{(7 \pm 1)}{2}}\)  →  \(\color{blue}{x = 4}\) or \(\color{blue}{x = 3}\).
Solutions: \(\color{blue}{x = 4}\) and \(\color{blue}{x = 3}\).

Example 4 (Quadratic Formula): Solve \(\color{blue}{2x^{2} + 5x – 3 = 0}\).

\(\color{blue}{a=2, b=5, c=-3}\). Discriminant: \(\color{blue}{25 + 24 = 49}\).
\(\color{blue}{x = \frac{(-5 \pm 7)}{4}}\)  →  \(\color{blue}{x = \frac{1}{2}}\) or \(\color{blue}{x = -3}\).
Solutions: \(\color{blue}{x = \frac{1}{2}}\) and \(\color{blue}{x = -3}\).

More Practice: Using the Quadratic Formula

Khan Academy walks through applying the quadratic formula step by step:

Exercises for Solving Quadratic Equations

1. Solve by factoring: \(\color{blue}{x^{2} – 7x + 12 = 0}\)
2. Solve by factoring: \(\color{blue}{x^{2} + x – 6 = 0}\)
3. Solve using square roots: \(\color{blue}{x^{2} = 81}\)
4. Solve using square roots: \(\color{blue}{3x^{2} – 48 = 0}\)
5. Solve using the quadratic formula: \(\color{blue}{x^{2} + 4x – 5 = 0}\)
6. Solve using the quadratic formula: \(\color{blue}{2x^{2} – 3x – 2 = 0}\)

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Answers

1. \(\color{blue}{x = 3}\) and \(\color{blue}{x = 4}\)
2. \(\color{blue}{x = 2}\) and \(\color{blue}{x = -3}\)
3. \(\color{blue}{x = \pm 9}\)
4. \(\color{blue}{x = \pm 4}\)
5. \(\color{blue}{x = 1}\) and \(\color{blue}{x = -5}\)
6. \(\color{blue}{x = 2}\) and \(\color{blue}{x = -\frac{1}{2}}\)

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Free Solving a Quadratic Equation Worksheet

Ready to practice on your own? Download our free Solving a Quadratic Equation worksheet below, work through each problem at your own pace, and then check your answers. If a few give you trouble, scroll back up to the worked examples and try again — steady practice is the surest way to master Solving a Quadratic Equation before a quiz or test.

Download The Quadratic Formula Worksheet

Frequently Asked Questions

Which method should I use to solve a quadratic?

Start with factoring — it is the fastest. If the equation doesn’t factor nicely (coefficients that lead to large numbers or irrational roots), use the quadratic formula. Use the square root method only when there is no \(\color{blue}{x}\) term (only \(\color{blue}{x^{2}}\) and a constant).

What does it mean when a quadratic has no real solutions?

If the discriminant \(\color{blue}{b^{2} – 4\text{ ac } < 0}\), the square root is imaginary and there are no real solutions. The parabola does not cross the x-axis.

Can a quadratic have exactly one solution?

Yes — when the discriminant equals zero, both solutions from the quadratic formula are the same. This is a double root and means the parabola just touches the x-axis at its vertex.

Related Topics

• Quadratic Function
• Factoring Quadratics
• Completing the Square
• Discriminant of a Quadratic Equation
• Graphing Quadratic Functions