Algebra 2 (1st Edition)

The solutions are $-2+\displaystyle \frac{\sqrt{26}}{2}$ and $-2-\displaystyle \frac{\sqrt{26}}{2}$.
$2(x+2)^{2}-5=8\qquad$ ...add $5$ to each side. $2(x+2)^{2}-5+5=8+5\qquad$ ...simplify. $2(x+2)^{2}=13\qquad$ ...divide each side with $2$. $(x+2)^{2}=\displaystyle \frac{13}{2}\qquad$ ...take square roots of each side. $\sqrt{(x+2)^{2}}=\sqrt{\frac{13}{2}}\qquad$ ...simplify. ...When solving an equation of the form $x^{2}=s$ where $s>0$, we find both the positive and negative solutions. $x+2=\pm\sqrt{\frac{13}{2}}\qquad$ ...apply the Quotient Property:$\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $x+2=\displaystyle \pm\frac{\sqrt{13}}{\sqrt{2}}\qquad$ ...rationalize the denominator by multiplying both the numerator and the denominator with $\sqrt{2}$ $x+2=\displaystyle \pm\frac{\sqrt{13}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $x+2=\displaystyle \pm\frac{\sqrt{26}}{\sqrt{4}}\qquad$ ...evaluate $\sqrt{4}$ ($\sqrt{4}=2$) $x+2=\displaystyle \pm\frac{\sqrt{26}}{2}\qquad$ ...add $-2$ to each side. $x+2-2=\displaystyle \pm\frac{\sqrt{26}}{2}-2\qquad$ ...simplify. $x=-2\displaystyle \pm\frac{\sqrt{26}}{2}$