# How do you solve 8p ^ { 2} + 38p + 62= - 8p + 6?

Oct 22, 2017

$p = - 1.75 \mathmr{and} p = - 4$

#### Explanation:

1) Rearrange so that one side = 0:

$8 {p}^{2} + 38 p + 62 + 8 p - 6 = - 8 p + 6 + 8 p - 6$
(Adding $8 p - 6$ to both sides)

$8 {p}^{2} + 46 p + 56 = 0$

$p = \frac{- b \pm \left(\sqrt{{b}^{2} - 4 a c}\right)}{2 a}$

Where a = 8, b = 46, c = 56.

$p = \frac{- 46 \pm \left(\sqrt{{46}^{2} - 4 \cdot 8 \cdot 56}\right)}{2 \cdot 8}$

$p = \frac{- 46 \pm \sqrt{324}}{16}$

$p = - 1.75 \mathmr{and} p = - 4$

Oct 22, 2017

$p = - 4$ or $p = \frac{- 7}{4}$

#### Explanation:

$8 {p}^{2} + 38 p + 62 = - 8 p + 6$

Subtract $\left(- 8 p + 6\right)$ from both sides.

$8 {p}^{2} + 38 p + 62 - \textcolor{red}{\left(- 8 p + 6\right)} = - 8 p + 6 - \textcolor{red}{\left(- 8 p + 6\right)}$

$8 {p}^{2} + 46 p + 56 = 0$

Divide both sides by $2$

$\frac{8 {p}^{2}}{\textcolor{red}{2}} + \frac{46 p}{\textcolor{red}{2}} + \frac{58}{\textcolor{red}{2}} = \frac{0}{\textcolor{red}{2}}$

$4 {p}^{2} + 23 p + 28 = 0$

Now we have to find two numbers whose product is equal to $28 \times 4$ ($28 \times 4 = 112$) and add up to $23$.

The two numbers are $16$ and $7$

So $4 {p}^{2} + 23 p + 28 = 0$ can be written as $4 {p}^{2} + 16 p + 7 p + 28 = 0$

$\left(4 {p}^{2} + 16 p\right) + \left(7 p + 28\right) = 0$

$4 p \left(p + 4\right) + 7 \left(p + 4\right) = 0$

$\left(p + 4\right) \left(4 p + 7\right) = 0$

So $\left(p + 4\right) = 0$ and $\left(4 p + 7\right) = 0$

$\textcolor{red}{1.}$$p + 4 = 0$

$p = - 4$

and

$\textcolor{red}{2.}$$4 p + 7 = 0$

$4 p = - 7$

$p = \frac{- 7}{4}$