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 - 9-6 The Quadratic Formula and the Discriminant - Practice and Problem-Solving Exercises - Page 586: 8


$x=-6$ $x={\dfrac{14}{5}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $5x^{2}+16x-84=0$.Therefore, $a=5$,$b=16$ and $c=-84$: Substitute the values: $x=\dfrac{-16±{\sqrt {16^{2}-4(5)(-84)}}}{2(5)}$ =$\dfrac{-16±{\sqrt {256-(-1680)}}}{10}$ =$\dfrac{-16± {\sqrt {1936}}}{10}$ =$\dfrac{-16±44}{10}$ Separate the equation into plus and minus equations: $x=\dfrac{-16-44}{10}=-6$ $x=\dfrac{-16+44}{10}={\dfrac{14}{5}}$
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