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: 12

Answer

{$-7,4$}

Work Step by Step

Using the rules of factoring trinomials, we obtain: $n^{2}+3n-28=0$ $n^{2}-4n+7n-28=0$ $n(n-4)+7(n-4)=0$ $(n-4)(n+7)=0$ $(n-4)=0$ and $(n+7)=0$ $n=4$ and $n=-7$ Therefore, the solution set is {$-7,4$}.
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