## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 25

#### 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 $t=-\displaystyle \frac{8}{3}$

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