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Quadratic Equations: Every Way to Solve One, Explained

October 2, 2026 · AI Math Tutor Team

A quadratic equation is any equation that can be written in the standard form $ax^2 + bx + c = 0$ , where a is not 0, and solving one means finding every value of x that makes the equation true. There are four standard ways to solve a quadratic equation: factoring, the square root method, completing the square, and the quadratic formula. Each one fits a different kind of quadratic equation best, and this guide walks through all four, with a table to help you pick the right one and one worked example for each method.

Key takeaways

A quadratic equation has $x^2$ as its highest power and can be written in standard form as $ax^2 + bx + c = 0$ .
Factoring is usually the fastest method when the equation breaks into two simple binomials.
The square root method works best when there is no x term, only $x^2$ and a constant.
Completing the square rewrites the equation as a squared binomial and works on any quadratic equation.
The quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , solves every quadratic equation, which makes it the reliable method when the others do not fit cleanly.

What makes an equation quadratic

An equation is quadratic when its highest power of the variable is 2, written in standard form as $ax^2 + bx + c = 0$ . The number a is the coefficient of $x^2$ and cannot be 0, or the equation would no longer have a squared term at all. The number b is the coefficient of x, and c is the constant term; either one can be 0 without changing what kind of equation it is.

A quadratic equation graphs as a parabola, a symmetric U-shaped curve, and the solutions to the equation are exactly the x-values where that curve crosses the x-axis. A parabola can cross the x-axis twice, once, or not at all, which is why a quadratic equation can have two solutions, one repeated solution, or no real solutions at all.

Before you solve, it helps to identify a, b, and c directly from the equation. In $2x^2 - 5x + 3 = 0$ , $a = 2$ , $b = -5$ , and $c = 3$ . If the equation is not already set equal to 0, move every term to one side first: $x^2 + 4x = 12$ becomes $x^2 + 4x - 12 = 0$ by subtracting 12 from both sides, which puts it in the standard form every method in this guide assumes.

Why a quadratic can have two, one, or no solutions

The number of real solutions a quadratic equation has depends on the discriminant, the expression $b^2 - 4ac$ found under the square root in the quadratic formula. When $b^2 - 4ac$ is positive, the equation has two distinct real solutions, since the square root of a positive number gives two possible signs. When $b^2 - 4ac$ equals 0, the equation has exactly one real solution, sometimes called a repeated or double solution, because adding or subtracting a square root of 0 gives the same value either way.

When $b^2 - 4ac$ is negative, the equation has no real solutions at all, since there is no real number whose square is negative. On the graph, this means the parabola never touches the x-axis; it stays entirely above it or entirely below it. Checking the discriminant before you start solving can save time, since it tells you what kind of answer to expect from any of the four methods below.

Four ways to solve a quadratic equation

Method	When to use it	Example
Factoring	The equation factors into two simple binomials	$x^2 + 5x + 6 = 0$ becomes $(x+2)(x+3) = 0$
Square root method	There is no x term, only $x^2$ and a constant	$x^2 = 25$ gives $x = \pm 5$
Completing the square	Factoring is not clean and you want the vertex form	$x^2 + 6x + 5 = 0$ becomes $(x+3)^2 = 4$
Quadratic formula	Works on any quadratic equation	$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Solving by factoring

Factoring rewrites $ax^2 + bx + c$ as a product of two binomials, then uses the fact that if two things multiply to 0, at least one of them has to be 0. For $x^2 + bx + c = 0$ with a leading coefficient of 1, look for two numbers that multiply to c and add to b. Those two numbers become the constants in the two binomials.

Once the equation is factored into $(x + p)(x + q) = 0$ , set each factor equal to 0 separately and solve for x. Factoring is usually the quickest method when the numbers work out to whole numbers, but not every quadratic factors that way. When the leading coefficient a is not 1, the process needs one extra step: multiply a and c together first, find two numbers that multiply to that product and add to b, then rewrite and factor by grouping. For the full pattern, including that extra step, see Quadratic Factoring: How to Factor and Solve.

Solving by the square root method

The square root method is the fastest option when the equation has no x term at all, just $x^2$ and a constant, such as $x^2 = 25$ or $3x^2 - 27 = 0$ . Isolate $x^2$ on one side of the equation, then take the square root of both sides, remembering to include both the positive and negative root.

That plus-or-minus step matters because squaring a positive number and squaring its negative both give the same positive result, so both roots make the original equation true. If the number under the square root is negative, such as in $x^2 = -9$ , the equation has no real solutions, since no real number squares to give a negative result. For more worked practice with this method, see Quadratic Equations by the Square Root Method.

Solving by completing the square

Completing the square turns $x^2 + bx + c = 0$ into a perfect square binomial by adding a carefully chosen number to both sides. Move the constant term to the other side of the equation first, then take half of b, square it, and add that number to both sides. The left side is now a perfect square trinomial, which factors into a squared binomial, and you can finish by taking the square root of both sides just like the square root method.

This method is more work than factoring when factoring is available, but it always works, and it also produces the vertex form of a parabola along the way, which shows the highest or lowest point of the parabola directly. The vertex form, $y = a(x - h)^2 + k$ , names the vertex directly as the point $(h, k)$ , which is often what the square root and factoring methods leave hidden.

One detail that trips people up: when a is not 1, divide every term by a before completing the square, so the coefficient of $x^2$ is 1 again. Skipping that step and completing the square on $2x^2 + 8x + 3 = 0$ directly, without dividing by 2 first, leads to the wrong number being added to both sides.

Solving with the quadratic formula

The quadratic formula solves any equation in standard form, no matter whether it factors nicely or not:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Identify a, b, and c from the equation's standard form, substitute them into the formula, and simplify. Keep track of the order of operations carefully: square b first, then subtract $4ac$ , then take the square root of the whole result before dividing by $2a$ . The formula works whether a, b, or c is positive, negative, a fraction, or even 0 for b or c, which is exactly what makes it the one method that never fails to apply.

Worked examples

Example 1: solve by factoring, $x^2 + 5x + 6 = 0$

Find two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3, since $2 \times 3 = 6$ and $2 + 3 = 5$ .
Write the factored form using those two numbers. $x^2 + 5x + 6 = (x+2)(x+3) = 0$
Set each factor equal to 0 and solve. $x + 2 = 0 \Rightarrow x = -2, \quad x + 3 = 0 \Rightarrow x = -3$

So $x = -2$ or $x = -3$ . Check by substituting back: $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$ , and $(-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0$ , both correct.

Example 2: solve by the square root method, $2x^2 - 8 = 0$

Isolate $x^2$ by adding 8 to both sides and dividing by 2. $2x^2 = 8 \Rightarrow x^2 = 4$
Take the square root of both sides, keeping both the positive and negative root. $x = \pm\sqrt{4} = \pm 2$

So $x = 2$ or $x = -2$ . Check by substituting back: $2(2)^2 - 8 = 8 - 8 = 0$ , and $2(-2)^2 - 8 = 8 - 8 = 0$ , both correct.

Example 3: solve by completing the square, $x^2 + 6x + 5 = 0$

Move the constant term to the other side of the equation. $x^2 + 6x = -5$
Take half of the coefficient of x, which is 6, square it to get 9, and add it to both sides. $x^2 + 6x + 9 = -5 + 9 = 4$
Write the left side as a squared binomial and take the square root of both sides. $(x+3)^2 = 4 \Rightarrow x + 3 = \pm 2$
Solve for x. $x = -3 + 2 = -1, \quad x = -3 - 2 = -5$

So $x = -1$ or $x = -5$ . Check by substituting back: $(-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0$ , and $(-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0$ , both correct.

Example 4: solve with the quadratic formula, $2x^2 + 3x - 2 = 0$

Identify a, b, and c from the standard form: $a = 2$ , $b = 3$ , $c = -2$ .
Substitute into the quadratic formula. $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}$
Compute both solutions separately. $x = \frac{2}{4} = \frac{1}{2}, \quad x = \frac{-8}{4} = -2$

So $x = \frac{1}{2}$ or $x = -2$ . Check by substituting back: $2\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) - 2 = 0.5 + 1.5 - 2 = 0$ , and $2(-2)^2 + 3(-2) - 2 = 8 - 6 - 2 = 0$ , both correct.

Common mistakes

Forgetting the plus-or-minus sign when taking a square root, which loses one of the two solutions.
Trying to factor a quadratic that does not factor over whole numbers instead of switching to the quadratic formula.
Making an arithmetic slip inside the square root of the quadratic formula, especially with a negative b or a negative $4ac$ .
Forgetting to divide every term by a, not just some of them, when isolating $x^2$ or completing the square.
Stopping after finding one solution, when a quadratic equation almost always has two.

Factoring is usually the fastest method when it works, and Quadratic Factoring: How to Factor and Solve covers the pattern in more depth, including quadratics with a leading coefficient other than 1 and quadratics that do not factor at all. For more practice with the factoring method specifically, see Solve a Quadratic Equation by Factoring: 5 Worked Examples. If your equation has no middle term at all, Quadratic Equations by the Square Root Method is the fastest path to the answer. And for a method that extends beyond quadratics to any kind of math problem, see How to Solve Any Math Problem Step by Step. When you want to check which method fits your equation, scan it and read every step in AI Math Tutor.

AI Math Tutor is a study aid, not a substitute for a teacher, and it can make mistakes: check every step, not just the answer.

Questions

What makes an equation a quadratic equation?

A quadratic equation has a variable raised to the second power as its highest power, and it can be written in standard form as $ax^2 + bx + c = 0$ , where a, b, and c are numbers and a is not 0.

Which method should you use to solve a quadratic equation?

Try factoring first if the numbers look simple, since it is usually the fastest method. Use the square root method if there is no x term at all. Use completing the square if you need the vertex form. The quadratic formula works on every quadratic equation, so it is the reliable fallback when the others do not fit.

Why does a quadratic equation usually have two solutions?

A quadratic equation graphs as a parabola, a U-shaped curve, which can cross the x-axis at up to two points. Each crossing point is a solution, which is why methods like the square root method and the quadratic formula both produce a plus-or-minus result.

What does it mean if the number under the square root in the quadratic formula is negative?

It means the equation has no real number solutions, because you cannot take the square root of a negative number and get a real number back. The parabola for that equation never crosses the x-axis.

Do you always need the quadratic formula?

No. The quadratic formula always works, but factoring or the square root method are often faster when the equation is set up for them. The quadratic formula is most useful when those other methods do not apply cleanly.