Answer
{-8, 4}
Work Step by Step
$x^2+4x-32 = 0$
$x^2+4x+4-36=0$
$(x+2)^2-6^2 = (x+2-6)(x+2+6)=(x-4)(x+8)$ (as $x^2-y^2=(x-y)(x+y)$
Hence either (x-4)=0 that gives x=4
or (x+8)=0 that gives x=-8
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 110: 33
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