Answer
The solutions are $2\sqrt{35}$ and $-2\sqrt{35}.$
Work Step by Step
$\displaystyle \frac{t^{2}}{20}+8=15\qquad$ ...add $-8$ to each side.
$\displaystyle \frac{t^{2}}{20}=7\qquad$ ...multiply each side with $20$.
$ t^{2}=140\qquad$ ...take square roots of each side.
$\sqrt{t^{2}}=\sqrt{140}\qquad$ ...simplify.
...When solving an equation of the form $x^{2}=s$ where $s>0$,
we find both the positive and negative solutions.
$ t=\pm\sqrt{140}\qquad$ ...rewrite $140$ as a product of two factors so that one factor is a perfect square. ($140=4\cdot 35$)
$ t=\pm\sqrt{4}\cdot 35\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$
$ t=\pm\sqrt{4}\cdot\sqrt{35}\qquad$ ...evaluate $\sqrt{4} $ ($\sqrt{4}=2$)
$t=\pm 2\sqrt{35}$
