Answer
$t=\displaystyle \frac{1}{5}$ or $t=-\displaystyle \frac{2}{3}$
Work Step by Step
$ 15t^{2}+7t=2\qquad$...add $-2$ to both sides.
$ 15t^{2}+7t-2=0\qquad$... solve with the Quadractic formula. $a=15,\ b=7,\ c=-2$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $7,\ a$ for $15$ and $c$ for $-2$.
$ t=\displaystyle \frac{-7\pm\sqrt{(7)^{2}-4\cdot(-2)\cdot 15}}{2\cdot 15}\qquad$... simplify.
$t=\displaystyle \frac{-7\pm\sqrt{49+120}}{30}$
$t=\displaystyle \frac{-7\pm\sqrt{169}}{30}$
$ t=\displaystyle \frac{-7\pm 13}{30}\qquad$... the symbol $\pm$ indicates two solutions.
$t=\displaystyle \frac{-7+13}{30}$ or $t=\displaystyle \frac{-7-13}{30}$
$t=\displaystyle \frac{6}{30}$ or $t=-\displaystyle \frac{20}{30}$
$t=\displaystyle \frac{1}{5}$ or $t=-\displaystyle \frac{2}{3}$
