Solve Using the Quadratic Formula 16y^2-8y=0

$16 y^{2} - 8 y = 0$

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 $a = 16$ , $b = - 8$ , and $c = 0$ into the quadratic formula and solve for $y$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (16 \cdot 0)}}{2 \cdot 16}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$y = \frac{8 \pm \sqrt{64 - 4 \cdot 16 \cdot 0}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot 0$ .

Multiply $- 4$ by $16$ .

$y = \frac{8 \pm \sqrt{64 - 64 \cdot 0}}{2 \cdot 16}$

Multiply $- 64$ by $0$ .

$y = \frac{8 \pm \sqrt{64 + 0}}{2 \cdot 16}$

Add $64$ and $0$ .

$y = \frac{8 \pm \sqrt{64}}{2 \cdot 16}$

Rewrite $64$ as $8^{2}$ .

$y = \frac{8 \pm \sqrt{8^{2}}}{2 \cdot 16}$

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

$y = \frac{8 \pm 8}{2 \cdot 16}$

Multiply $2$ by $16$ .

$y = \frac{8 \pm 8}{32}$

Simplify $\frac{8 \pm 8}{32}$ .

$y = \frac{1 \pm 1}{4}$

Step 4

The final answer is the combination of both solutions.

$y = \frac{1}{2}, 0$