Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.1 - Quadratic Equations - Problem Set 10.1 - Page 442: 13



Work Step by Step

Using the rules of factoring trinomials, we obtain: $6y^{2}+7y-5=0$ $6y^{2}-3y+10y-5=0$ $3y(2y-1)+5(2y-1)=0$ $(2y-1)(3y+5)=0$ $(2y-1)=0$ and $(3y+5)=0$ $y=\frac{1}{2}$ and $y=-\frac{5}{3}$ Therefore, the solution set is {$-\frac{5}{3},\frac{1}{2}$}.
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