# How do you solve for x in simplest radical form: 2(x+3)^2+10=66?

Apr 14, 2015

$2 {\left(x + 3\right)}^{2} + 10 = 66$

${\left(x + 3\right)}^{2} + 5 = 33$

${x}^{2} + 6 x + 9 + 5 - 33 = 0$

${x}^{2} + 6 x - 19 = 0$

Using the quadratic root formula: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$

$x = \frac{- 6 \pm \sqrt{36 + 76}}{2}$

$x = \frac{- 6 \pm 4 \sqrt{7}}{2}$

$x = - 3 - 2 \sqrt{7}$
or
$x = - 3 + 2 \sqrt{7}$

Apr 14, 2015

Alternative solution:

$2 {\left(x + 3\right)}^{2} + 10 = 66$

$2 {\left(x + 3\right)}^{2} = 66 - 10$ $\textcolor{w h i t e}{\text{ss}}$ added $- 10$ on both sides

$2 {\left(x + 3\right)}^{2} = 56$

${\left(x + 3\right)}^{2} = \frac{56}{2}$$\textcolor{w h i t e}{\text{ssssssss}}$ multiplied $\frac{1}{2}$ on both sides

${\left(x + 3\right)}^{2} = 28$

$x + 3 = \pm \sqrt{28}$ $\textcolor{w h i t e}{\text{ssssss}}$ there are 2 numbers whise square is 28

$x = - 3 \pm \sqrt{28}$ $\textcolor{w h i t e}{\text{ss}}$ added $- 3$ on both sides

$x = - 3 \pm \sqrt{4 \cdot 7} = - 3 \pm \sqrt{4} \sqrt{7} = - 3 \pm 2 \sqrt{7}$