Answer
y=$\frac{1}{4}$ or $y=-\frac{3}{5}$
Work Step by Step
$20y^{2}+7y+3$
=$20y^{2}+12y-5y+3$
=$4y(5y+3)-1(5y+3)$
$(4y-1)(5y+3)$$=0$
y=$\frac{1}{4}$ or $y=-\frac{3}{5}$
Finite Math and Applied Calculus (6th Edition)
by Waner, Stefan; Costenoble, Steven
Published by
Brooks Cole
ISBN 10:
1133607705
ISBN 13:
978-1-13360-770-0
Chapter 0 - Section 0.3 - Multiplying and Factoring Algebraic Expressions - Exercises - Page 21: 44
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