Consider a quadratic equation in standard form:

Solving the quadratic equation means finding the values of x, called roots, that make this equation true (i.e. makes the left hand side equal to 0.)

The formula below will find the two roots of the equation:

There are two roots because the ± symbol means consider it as a + one time and as a another time.

# 1

Compare the quadratic equation in the question with the standard form.

2x2 −5x + 2 = 0 ⇔ ax2 + bx + c = 0

Find the values of a, b and c.

a = 2, b = −5, c = 2

# 2

Use the formula.

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

# 3

Substitute a, b and c into the formula. In our example, a = 2, b = −5 and c = 2.

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}$$ $$\:\:\:\: = \frac{5 \pm \sqrt{(-5 \times -5) - 4 \times 2 \times 2}}{2 \times 2}$$ $$\:\:\:\: = \frac{5 \pm \sqrt{25 - 16}}{4}$$ $$\:\:\:\: = \frac{5 \pm \sqrt{9}}{4}$$ $$\:\:\:\: = \frac{5 \pm 3}{4}$$

# 4

Find the root that comes from turning the ± into a +.

$$x = \frac{5 + 3}{4}$$ $$\:\:\:\: = \frac{8}{4}$$ $$\:\:\:\: = 8 \div 4$$ $$\mathbf{x = 2}$$

x = 2 is a root of the quadratic equation.

# 5

Find the root that comes from turning the ± into a −.

$$x = \frac{5 - 3}{4}$$ $$\:\:\:\: = \frac{2}{4}$$ $$\mathbf{x = \frac{1}{2}}$$

x = 12 is a root of the quadratic equation.

We have solved the quadratic equation:

x = 12, x = 2.

## Slider

Sometimes quadratic equations have repeated roots: the same value of x solves the quadratic equation twice.

The slider below shows another real example of how to solve a quadratic equation using the quadratic formula.

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