Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4 - Page 457: 18



Work Step by Step

Factoring the equation and setting the factors equal to zero, we obtain: $20y^{2}-7y-6=0$ $20y^{2}+8y-15y-6=0$ $4y(5y+2)-3(5y+2)=0$ $(5y+2)(4y-3)=0$ $(5y+2)=0$ or $(4y-3)=0$ $y=-\frac{2}{5}$ or $y=\frac{3}{4}$ Therefore, the solution set is {$-\frac{2}{5},\frac{3}{4}$}.
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.