Quadratic equations, and how to solve them

Updated October 1, 2026

A quadratic equation is an equation where the highest power of the variable is 2, written in standard form as ax² + bx + c = 0 with a not equal to 0. It can have two, one, or no real solutions. Students solve them by factoring, taking square roots, completing the square, or using the quadratic formula, all taught in Algebra I or Integrated Math II in California.

What makes an equation quadratic

An equation is quadratic when the variable is squared and no higher power appears. In standard form, ax² + bx + c = 0, the numbers a, b, and c are called coefficients. The a cannot be zero, or the squared term disappears and you are back to a linear equation.

The graph of y = ax² + bx + c is a U-shaped curve called a parabola. The solutions of ax² + bx + c = 0 are the x-values where the parabola crosses the x-axis, which is why solutions are also called roots or zeros. A parabola can cross the x-axis twice, touch it once, or miss it entirely, which matches having two, one, or no real solutions.

In the California Common Core State Standards for Mathematics, solving quadratic equations in one variable, and factoring a quadratic to reveal its zeros, are Algebra I standards. Solutions that are not real numbers, written with i, are handled later in Algebra II.

Solving by factoring

Factoring works when the quadratic splits neatly into two binomials. It rests on one fact: if two numbers multiply to zero, at least one of them must be zero. For x² + bx + c with a = 1, look for two numbers that multiply to c and add to b.

Problem $x^{2} - 5 x + 6 = 0$

$(- 2) (- 3)$ $= 6,$ $(- 2) + (- 3)$ $= - 5$
Find two numbers that multiply to 6 and add to −5.
$(x - 2) (x - 3)$ $= 0$
Use them to write the factored form.
$x - 2$ $= 0$ $or$ $x - 3$ $= 0$
If a product is zero, one of the factors is zero.
$x$ $= 2$ $or$ $x$ $= 3$
Solve each small equation.
$2^{2} - 5 (2) + 6$ $= 0,$ $3^{2} - 5 (3) + 6$ $= 0$
Check both answers in the original equation.

Answer $x = 2 or x = 3$

Solving with the quadratic formula

Many quadratics do not factor with whole numbers. The quadratic formula works for every quadratic in standard form. It comes from completing the square on ax² + bx + c = 0, which is why students are often asked to derive it.

Problem $2 x^{2} + 3 x - 4 = 0$

$a$ $= 2,$ $b$ $= 3,$ $c$ $= - 4$
Read the coefficients from standard form, keeping the sign on c.
$x$ $= \frac{- b \pm \sqrt b^{2} - 4 a c}{2 a}$
Write the quadratic formula.
$b^{2} - 4 a c$ $= 3^{2} - 4 (2) (- 4)$ $= 9 + 32$ $= 41$
Find the discriminant first. Minus a negative becomes plus.
$x$ $= \frac{- 3 \pm \sqrt 41}{4}$
Substitute into the formula. The denominator is 2a, which is 4.
$x$ $\approx 0.851$ $or$ $x$ $\approx - 2.351$
Estimate the two solutions to check that they are reasonable.

Answer $x = \frac{- 3 \pm \sqrt 41}{4}$

Other methods and how to choose

Square roots: when there is no bx term, or the equation is already a perfect square, isolate the square and take the square root of both sides. For (x − 1)^2 = 16, x − 1 is 4 or −4, so x = 5 or x = −3. Remember both signs.

Completing the square: rewrite the quadratic so one side is a perfect square. For x² + 4x − 1 = 0, add 1 to get x² + 4x = 1, then add 4 to both sides to get (x + 2)^2 = 5, so x = −2 plus or minus the square root of 5.

Completing the square is also how you find the vertex. For example, x² − 6x + 5 can be written as (x − 3)^2 − 4, so its parabola has its lowest point at (3, −4).

Try factoring first when the numbers are small and a is 1.
Use square roots when there is no bx term.
Use the quadratic formula when factoring is not obvious. It always works.
Use completing the square when you need the vertex form of the parabola.

What the discriminant tells you

The expression under the square root, b² − 4ac, is called the discriminant. You can compute it before solving to know what kind of answer to expect.

If it is positive, there are two real solutions, as in the 2x² + 3x − 4 example where it was 41. If it is zero, there is one repeated solution. For x² + 4x + 4 = 0, the discriminant is 16 − 16 = 0 and the only solution is x = −2.

If it is negative, there are no real solutions. For x² + 2x + 5 = 0, the discriminant is 4 − 20 = −16, and the parabola never touches the x-axis. In Algebra II, students write those solutions as complex numbers, −1 + 2i and −1 − 2i.

Common mistakes with quadratics

Forgetting to move everything to one side before factoring. x² − 5x = −6 has to become x² − 5x + 6 = 0 first.
Losing the negative sign on c when substituting into the formula, which changes the discriminant.
Dividing by 2 instead of by 2a, or dividing only the square root part instead of the whole numerator.
Taking only the positive square root and missing the second solution.
Dividing both sides by x, which throws away the solution x = 0.

Questions

What is the easiest way to solve a quadratic equation?

Factoring is usually quickest when the numbers are small and the quadratic splits into whole-number factors. When it does not, the quadratic formula always works.

Can a quadratic equation have no solution?

It can have no real solution. That happens when the discriminant b² − 4ac is negative, so the parabola never crosses the x-axis. In Algebra II, those equations are solved with complex numbers.

Why do quadratic equations have two answers?

Squaring hides the sign, so two different numbers can produce the same result. Graphically, a parabola can cross a horizontal line at two points.

When do California students learn quadratic equations?

Quadratic equations and functions are a major part of Algebra I, or of Integrated Math II in the integrated sequence. Students return to them in Algebra II, where they also solve quadratics with complex solutions.

Do I need to memorize the quadratic formula?

Many classes expect students to know it, and it is worth memorizing because it works for every quadratic. Understanding where it comes from, by completing the square, makes it easier to remember.