Answer
$t=\displaystyle \frac{5}{4}$ or $t=-\displaystyle \frac{8}{3}$
Work Step by Step
$ 12t^{2}+17t=40\qquad$...add $-40$ to both sides.
$ 12t^{2}+17t-40=0\qquad$... solve with the Quadractic formula. $a=12,\ b=17,\ c=-40$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $17,\ a$ for $12$ and $c$ for $-40$.
$ t=\displaystyle \frac{-17\pm\sqrt{(17)^{2}-4\cdot(-40)\cdot 12}}{2\cdot 12}\qquad$... simplify.
$t=\displaystyle \frac{-17\pm\sqrt{289+1920}}{24}$
$t=\displaystyle \frac{-17\pm\sqrt{2209}}{24}$
$ t=\displaystyle \frac{-17\pm 47}{24}\qquad$... the symbol $\pm$ indicates two solutions.
$t=\displaystyle \frac{-17+47}{24}$ or $t=\displaystyle \frac{-17-47}{24}$
$t=\displaystyle \frac{30}{24}$ or $t=-\displaystyle \frac{64}{24}$
$t=\displaystyle \frac{5}{4}$ or $t=-\displaystyle \frac{8}{3}$
