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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 26


$t=\displaystyle \frac{1}{5}$ 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 $t=-\displaystyle \frac{2}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.