Answer
{$-6,1$}
Work Step by Step
Using the rules of factoring trinomials, we obtain:
$n^{2}+5n-6=0$
$n^{2}-1n+6n-6=0$
$n(n-1)+6(n-1)=0$
$(n-1)(n+6)=0$
$(n-1)=0$ and $(n+6)=0$
$n=1$ and $n=-6$
Therefore, the solution set is {$-6,1$}.
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: 11
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