Solve Using the Quadratic Formula 5x^2-180=0

$5 x^{2} - 180 = 0$

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 $a = 5$ , $b = 0$ , and $c = - 180$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (5 \cdot - 180)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 5 \cdot - 180}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 180$ .

Multiply $- 4$ by $5$ .

$x = \frac{0 \pm \sqrt{0 - 20 \cdot - 180}}{2 \cdot 5}$

Multiply $- 20$ by $- 180$ .

$x = \frac{0 \pm \sqrt{0 + 3600}}{2 \cdot 5}$

Add $0$ and $3600$ .

$x = \frac{0 \pm \sqrt{3600}}{2 \cdot 5}$

Rewrite $3600$ as $60^{2}$ .

$x = \frac{0 \pm \sqrt{60^{2}}}{2 \cdot 5}$

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

$x = \frac{0 \pm 60}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{0 \pm 60}{10}$

Simplify $\frac{0 \pm 60}{10}$ .

$x = \pm 6$

Step 4

The final answer is the combination of both solutions.

$x = 6, - 6$