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: 12


$x=-12$ $x={\dfrac{-8}{3}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $3x^{2}+44x+96=0$.Therefore, $a=3$,$b=44$ and $c=96$ Substitute the values: $x=\dfrac{-44±{\sqrt {44^{2}-4(3)(96)}}}{2(3)}$ =$\dfrac{-44±{\sqrt {1936-1152}}}{6}$ =$\dfrac{-44± {\sqrt {784}}}{6}$ =$\dfrac{-44±28}{6}$ Separate the equation into plus and minus equations: $x=\dfrac{-44-28}{6}=-12$ $x=\dfrac{-44+28}{6}={\dfrac{-8}{3}}$
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