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 {$-7,4$}.
Elementary Algebra
by Kaufmann, Jerome; Schwitters, Karen;
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: 12
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