Solve Using the Quadratic Formula 4n^2-81=0

$4 n^{2} - 81 = 0$

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 $a = 4$ , $b = 0$ , and $c = - 81$ into the quadratic formula and solve for $n$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (4 \cdot - 81)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

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

$n = \frac{0 \pm \sqrt{0 - 4 \cdot 4 \cdot - 81}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 81$ .

Multiply $- 4$ by $4$ .

$n = \frac{0 \pm \sqrt{0 - 16 \cdot - 81}}{2 \cdot 4}$

Multiply $- 16$ by $- 81$ .

$n = \frac{0 \pm \sqrt{0 + 1296}}{2 \cdot 4}$

Add $0$ and $1296$ .

$n = \frac{0 \pm \sqrt{1296}}{2 \cdot 4}$

Rewrite $1296$ as $36^{2}$ .

$n = \frac{0 \pm \sqrt{36^{2}}}{2 \cdot 4}$

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

$n = \frac{0 \pm 36}{2 \cdot 4}$

Multiply $2$ by $4$ .

$n = \frac{0 \pm 36}{8}$

Simplify $\frac{0 \pm 36}{8}$ .

$n = \frac{\pm 9}{2}$

Step 4

The final answer is the combination of both solutions.

$n = \frac{9}{2}, - \frac{9}{2}$