Solve Using the Quadratic Formula 9y^2-16=0

$9 y^{2} - 16 = 0$

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 $a = 9$ , $b = 0$ , and $c = - 16$ into the quadratic formula and solve for $y$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (9 \cdot - 16)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{0 \pm \sqrt{0 - 4 \cdot 9 \cdot - 16}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 16$ .

Multiply $- 4$ by $9$ .

$y = \frac{0 \pm \sqrt{0 - 36 \cdot - 16}}{2 \cdot 9}$

Multiply $- 36$ by $- 16$ .

$y = \frac{0 \pm \sqrt{0 + 576}}{2 \cdot 9}$

Add $0$ and $576$ .

$y = \frac{0 \pm \sqrt{576}}{2 \cdot 9}$

Rewrite $576$ as $24^{2}$ .

$y = \frac{0 \pm \sqrt{24^{2}}}{2 \cdot 9}$

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

$y = \frac{0 \pm 24}{2 \cdot 9}$

Multiply $2$ by $9$ .

$y = \frac{0 \pm 24}{18}$

Simplify $\frac{0 \pm 24}{18}$ .

$y = \frac{\pm 4}{3}$

Step 4

The final answer is the combination of both solutions.

$y = \frac{4}{3}, - \frac{4}{3}$