Solve Using the Quadratic Formula 2x^2-18x+40=0

$2 x^{2} - 18 x + 40 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 2$ , $b = - 18$ , and $c = 40$ into the quadratic formula and solve for $x$ .

$\frac{18 \pm \sqrt{{(- 18)}^{2} - 4 \cdot (2 \cdot 40)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 18$ to the power of $2$ .

$x = \frac{18 \pm \sqrt{324 - 4 \cdot 2 \cdot 40}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 40$ .

Multiply $- 4$ by $2$ .

$x = \frac{18 \pm \sqrt{324 - 8 \cdot 40}}{2 \cdot 2}$

Multiply $- 8$ by $40$ .

$x = \frac{18 \pm \sqrt{324 - 320}}{2 \cdot 2}$

Subtract $320$ from $324$ .

$x = \frac{18 \pm \sqrt{4}}{2 \cdot 2}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{18 \pm \sqrt{2^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{18 \pm 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{18 \pm 2}{4}$

Simplify $\frac{18 \pm 2}{4}$ .

$x = \frac{9 \pm 1}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 5, 4$