Answer
{$-7,5$}
Work Step by Step
Using the rules of factoring trinomials to factor, we obtain:
$a^{2}+2a-35=0$
$a^{2}+7a-5a-35=0$
$a(a+7)-5(a+7)=0$
$(a+7)(a-5)=0$
$(a+7)=0$ or $(a-5)=0$
$a=-7$ or $a=5$
Therefore, the solution set is {$-7,5$}.
Elementary Algebra
by Kaufmann, Jerome; Schwitters, Karen;
Published by
Cengage Learning
ISBN 10:
1285194055
ISBN 13:
978-1-28519-405-9
Chapter 11 - Additional Topics - 11.5 - Quadratic Equations: Complex Solutions - Problem Set 11.5 - Page 496: 12
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