How do you solve 4r ^ { 2} + 4r - 8= - 5?

Mar 10, 2018

$r = - \frac{3}{2} , \frac{1}{2}$

Explanation:

Solve:

$4 {r}^{2} + 4 r - 8 = - 5$

Add $5$ to both sides.

$4 {r}^{2} + 4 r - 8 + 5 = 0$

Simplify.

$4 {r}^{2} + 4 r - 3 = 0$ is a quadratic equation in standard form:

$a {x}^{2} + b x + c = 0 ,$

where:

$a = 4 ,$ $b = 4 ,$ $c = - 3$

You can solve this equation by using the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute $r$ for $x$, plug in the known values and solve.

$r = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

$r = \frac{- 4 \pm \sqrt{64}}{8}$

Simplify $\sqrt{64}$ to $8$.

$r = \frac{- 4 \pm 8}{8}$

$r = \frac{- 4 + 8}{8} ,$ $\frac{- 4 - 8}{8}$

$r = \frac{4}{8} ,$ $- \frac{12}{8}$

Reduce.

$r = \frac{1}{2} ,$ $- \frac{3}{2}$