Solve Using the Quadratic Formula 4b^2+8b=0

$4 b^{2} + 8 b = 0$

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 $a = 4$ , $b = 8$ , and $c = 0$ into the quadratic formula and solve for $b$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $0$ .

$b = \frac{- 8 \pm \sqrt{64 + 0}}{2 \cdot 4}$

Add $64$ and $0$ .

$b = \frac{- 8 \pm \sqrt{64}}{2 \cdot 4}$

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

$b = \frac{- 8 \pm \sqrt{8^{2}}}{2 \cdot 4}$

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

$b = \frac{- 8 \pm 8}{2 \cdot 4}$

Multiply $2$ by $4$ .

$b = \frac{- 8 \pm 8}{8}$

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

$b = - 1 \pm 1$

Step 4

The final answer is the combination of both solutions.

$b = 0, - 2$