Quadratic equations show up everywhere in algebra, from graphing parabolas to solving real-world problems about area, motion, and profit. If you have ever stared at an equation like \(x^2 + 5x + 6 = 0\) and wondered where to even begin, you are in the right place. The good news is that there are four reliable methods for solving these equations, and once you understand when to use each one, quadratics become far less intimidating.
In this guide you will learn exactly how to solve quadratic equations using four proven techniques: factoring, the square root property, completing the square, and the quadratic formula. We will work through every step of several examples, double-check each answer, and show you how to pick the fastest method for any problem. If you ever want to confirm your work or see a full step-by-step breakdown, you can always check it with our free equation solver.
By the end, you will have a clear strategy you can apply to homework, tests, and standardized exams. Let’s start with the basics so every method makes sense.
What Is a Quadratic Equation?
A quadratic equation is any equation that can be written in the standard form:
$$ax^2 + bx + c = 0$$Here, \(a\), \(b\), and \(c\) are numbers (called coefficients), and \(a \neq 0\). The defining feature is the \(x^2\) term, which is what makes the equation “quadratic” (from the Latin quadratus, meaning square). The solutions to a quadratic equation are called its roots or zeros, and most quadratics have two of them.
Key idea: identify a, b, and c first
Before choosing a method, rewrite your equation in standard form and write down the values of \(a\), \(b\), and \(c\). For \(2x^2 + 3x - 5 = 0\), we have \(a = 2\), \(b = 3\), and \(c = -5\). Getting these right is half the battle.
If your equation has terms on both sides, move everything to one side first so that it equals zero. For example, \(x^2 + 4 = 5x\) becomes \(x^2 - 5x + 4 = 0\). This single setup step is what every method below depends on.
The 4 Methods at a Glance
Each method has a sweet spot. The table below summarizes when to reach for each one, so you can save time and avoid unnecessary work.
| Method | Best used when… | Speed |
|---|---|---|
| Factoring | The expression factors nicely with whole numbers | Very fast |
| Square root property | There is no \(bx\) term (the form is \(ax^2 + c = 0\) or \((x-h)^2 = k\)) | Very fast |
| Completing the square | You need vertex form or the equation won’t factor cleanly | Moderate |
| Quadratic formula | Always works, especially for messy or irrational roots | Reliable |
A smart strategy is to try factoring first. If it doesn’t factor quickly, fall back on the quadratic formula, which never fails. Let’s see each method in action.
Method 1: Solving Quadratic Equations by Factoring
Factoring works because of the Zero Product Property: if two things multiply to zero, then at least one of them must be zero. So if we can rewrite \(ax^2 + bx + c = 0\) as a product of two binomials equal to zero, we can solve each piece separately.
Example 1: A simple trinomial
Solve \(x^2 + 5x + 6 = 0\).
Step 1 — Find two numbers that multiply to \(c = 6\) and add to \(b = 5\). Those numbers are \(2\) and \(3\), because \(2 \times 3 = 6\) and \(2 + 3 = 5\).
Step 2 — Write the factored form:
$$(x + 2)(x + 3) = 0$$Step 3 — Apply the Zero Product Property. Set each factor equal to zero:
$$x + 2 = 0 \quad \Rightarrow \quad x = -2$$ $$x + 3 = 0 \quad \Rightarrow \quad x = -3$$Step 4 — Check. For \(x = -2\): \((-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0\). For \(x = -3\): \((-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0\). Both work.
Example 2: When the leading coefficient isn’t 1
Solve \(2x^2 + 7x + 3 = 0\). Because \(a = 2\), we use the AC method (also called grouping).
Step 1 — Multiply \(a \times c\): \(2 \times 3 = 6\). Now find two numbers that multiply to \(6\) and add to \(b = 7\). Those numbers are \(1\) and \(6\).
Step 2 — Split the middle term using \(1\) and \(6\):
$$2x^2 + x + 6x + 3 = 0$$Step 3 — Group and factor:
$$x(2x + 1) + 3(2x + 1) = 0$$ $$(2x + 1)(x + 3) = 0$$Step 4 — Solve each factor:
$$2x + 1 = 0 \quad \Rightarrow \quad x = -\tfrac{1}{2}$$ $$x + 3 = 0 \quad \Rightarrow \quad x = -3$$Common mistake
Don’t forget to set each factor to zero. Students often factor correctly but then write only one solution. A quadratic usually has two roots, so check every factor.
Method 2: The Square Root Property
When a quadratic has no middle (\(bx\)) term, or when it is already written as a perfect square, the fastest approach is to take the square root of both sides. The crucial detail: a positive number has two square roots, one positive and one negative, so you must include the \(\pm\) symbol.
The square root property says: if \(x^2 = k\) and \(k \geq 0\), then \(x = \pm\sqrt{k}\). Forgetting the negative root is one of the most common errors in all of algebra.
Example 3: A perfect-square equation
Solve \(2(x - 3)^2 = 32\).
Step 1 — Isolate the squared term by dividing both sides by 2:
$$(x - 3)^2 = 16$$Step 2 — Take the square root of both sides, remembering the \(\pm\):
$$x - 3 = \pm\sqrt{16} = \pm 4$$Step 3 — Solve both cases:
$$x - 3 = 4 \quad \Rightarrow \quad x = 7$$ $$x - 3 = -4 \quad \Rightarrow \quad x = -1$$Step 4 — Check. For \(x = 7\): \(2(7 - 3)^2 = 2(16) = 32\). For \(x = -1\): \(2(-1 - 3)^2 = 2(16) = 32\). Both are correct.
Method 3: Completing the Square
Completing the square is a powerful technique that turns any quadratic into a perfect-square form so you can use the square root property. It is also the method used to derive the quadratic formula, and it is essential for finding the vertex of a parabola.
Example 4: Completing the square step by step
Solve \(x^2 + 6x + 5 = 0\).
Step 1 — Move the constant to the right side:
$$x^2 + 6x = -5$$Step 2 — Take half of the \(x\)-coefficient and square it. Half of \(6\) is \(3\), and \(3^2 = 9\). Add \(9\) to both sides:
$$x^2 + 6x + 9 = -5 + 9$$ $$x^2 + 6x + 9 = 4$$Step 3 — Factor the left side as a perfect square:
$$(x + 3)^2 = 4$$Step 4 — Apply the square root property:
$$x + 3 = \pm 2$$ $$x = -3 + 2 = -1 \quad \text{or} \quad x = -3 - 2 = -5$$Step 5 — Check. For \(x = -1\): \((-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0\). For \(x = -5\): \((-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0\). Both check out.
Watch out
If the leading coefficient \(a\) is not 1, divide every term by \(a\) first before completing the square. For \(2x^2 + 8x - 10 = 0\), divide by 2 to get \(x^2 + 4x - 5 = 0\), then continue as usual.
Method 4: The Quadratic Formula
The quadratic formula is the universal solver. It works for every quadratic equation, whether the roots are whole numbers, fractions, irrational numbers, or even complex numbers. Memorize it, and you will never be stuck:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$To use it, put your equation in standard form, identify \(a\), \(b\), and \(c\), and substitute carefully (especially with negative signs).
Example 5: Clean roots from the formula
Solve \(2x^2 + 3x - 5 = 0\). Here \(a = 2\), \(b = 3\), and \(c = -5\).
Step 1 — Compute the discriminant \(b^2 - 4ac\):
$$3^2 - 4(2)(-5) = 9 + 40 = 49$$Step 2 — Substitute into the formula:
$$x = \frac{-3 \pm \sqrt{49}}{2(2)} = \frac{-3 \pm 7}{4}$$Step 3 — Split into two solutions:
$$x = \frac{-3 + 7}{4} = \frac{4}{4} = 1$$ $$x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\tfrac{5}{2}$$Example 6: Irrational roots
Solve \(x^2 - 4x + 1 = 0\). Here \(a = 1\), \(b = -4\), and \(c = 1\). This equation does not factor with whole numbers, which is exactly when the formula shines.
Step 1 — Discriminant:
$$(-4)^2 - 4(1)(1) = 16 - 4 = 12$$Step 2 — Substitute:
$$x = \frac{-(-4) \pm \sqrt{12}}{2(1)} = \frac{4 \pm \sqrt{12}}{2}$$Step 3 — Simplify the radical. Since \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\):
$$x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$Study tip
Always simplify radicals and reduce fractions at the end. Leaving an answer as \(\frac{4 \pm 2\sqrt{3}}{2}\) is technically correct but usually marked incomplete. Reduce it to \(2 \pm \sqrt{3}\).
The Discriminant: Predicting Your Solutions
The expression under the square root, \(b^2 - 4ac\), is called the discriminant. Calculating it first tells you how many real solutions to expect, which can save you time and help you check your work.
| Discriminant \(b^2 - 4ac\) | Number of real solutions | What it means |
|---|---|---|
| Positive (\(> 0\)) | Two distinct real roots | Parabola crosses the x-axis twice |
| Zero (\(= 0\)) | One repeated real root | Parabola just touches the x-axis |
| Negative (\(< 0\)) | No real roots (two complex) | Parabola never touches the x-axis |
In Example 5, the discriminant was \(49\) (positive), so we correctly found two real roots. In Example 6 it was \(12\) (also positive), giving two irrational roots. If you compute a negative discriminant, the equation has no real solutions, only complex ones.
Try it instantly
Not sure whether your equation factors or which method is fastest? Enter it into our step-by-step equation solver and it will choose the best approach and show every line of work.
A Real-World Example: Using Quadratics to Solve Problems
Quadratic equations are not just abstract symbols; they model real situations. Here is a classic word problem.
Example 7: A rectangular garden
A rectangular garden has an area of 40 square feet. Its length is 3 feet more than its width. Find the dimensions.
Step 1 — Set up variables. Let the width be \(w\). Then the length is \(w + 3\), and area equals length times width:
$$w(w + 3) = 40$$Step 2 — Write in standard form:
$$w^2 + 3w - 40 = 0$$Step 3 — Factor. We need two numbers that multiply to \(-40\) and add to \(3\); those are \(8\) and \(-5\):
$$(w + 8)(w - 5) = 0$$Step 4 — Solve: \(w = -8\) or \(w = 5\). Since a width cannot be negative, we reject \(w = -8\). So the width is \(5\) feet and the length is \(5 + 3 = 8\) feet.
Step 5 — Check: \(5 \times 8 = 40\) square feet. Correct.
Notice how the negative solution was mathematically valid but physically impossible. Always check that your answers make sense in the context of the problem.
How to Choose the Right Method
With four tools in your kit, here is a quick decision guide:
- No middle term? (like \(x^2 - 9 = 0\)) Use the square root property.
- Factors easily with whole numbers? Use factoring; it’s the fastest.
- Need the vertex or working with a derivation? Use completing the square.
- Anything else, or roots look messy? Use the quadratic formula; it always works.
Remember that all four methods give the same correct answers; they are simply different routes to the same destination. Choosing wisely just saves you time.
Common Mistakes to Avoid
- Forgetting the \(\pm\): Taking a square root always produces two values. Missing the negative root loses half your answer.
- Sign errors with \(b\) and \(c\): When \(b\) or \(c\) is negative, substitute carefully. In Example 6, \(-(-4)\) became \(+4\).
- Not setting the equation to zero: Every method assumes standard form \(ax^2 + bx + c = 0\). Move all terms to one side first.
- Skipping the check: Plug your answers back in. It takes seconds and catches most errors.
Related Algebra Guides
Quadratics build on skills you already have and connect to topics you will study next. If you want to strengthen the foundations or branch out, these guides will help:
- How to Solve Linear Equations Step by Step — master single-variable equations before tackling quadratics.
- How to Solve Systems of Equations (Substitution, Elimination & Graphing) — combine multiple equations to find shared solutions.
- How to Solve Inequalities Step by Step — extend your skills to expressions with \(<\), \(>\), and \(\geq\).
Conclusion
You now have four dependable ways to solve any quadratic equation. Factoring is the quickest when the numbers cooperate, the square root property handles equations without a middle term, completing the square reveals structure and the vertex, and the quadratic formula works every single time. Pair these methods with the discriminant, and you can predict and verify your solutions with confidence.
The best way to get comfortable is to practice, then confirm your reasoning. Work through a few problems on your own, then drop them into our free equation solver to see each step laid out and to make sure your answers are right. For more clear, worked-example tutorials across algebra and beyond, visit the Math Solver AI blog. Keep practicing, and quadratics will soon feel routine.
