Elementary Algebra

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

Chapter 6 - Factoring, Solving Equations, and Problem Solving - Chapter 6 Review Problem Set - Page 274: 43



Work Step by Step

Using the rules of factoring trinomials to factor the polynomial, we obtain: $-7n-2n^{2}=-15$ $-2n^{2}-7n+15=0$ $2n^{2}+7n-15=0$ $2n^{2}-3n+10n-15=0$ $n(2n-3)+5(2n-3)=0$ $(2n-3)(n+5)=0$ Now, we equate the two factors to zero and solve: $(2n-3)(n+5)=0$ $(2n-3)=0$ or $(n+5)=0$ $n=\frac{3}{2}$ or $n=-5$ Therefore, the solution set is {$-5,\frac{3}{2}$}.
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