Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - Chapter Review - Page 605: 30


$x=\pm \sqrt 14 -3$

Work Step by Step

$x^2+6x-5=0$ $x^2+6x+9=14$ $(x+3)^2=14$ $x+3=\pm \sqrt 14$ $x=\pm \sqrt 14 -3$
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