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: 13


$x=-11$ $x={\dfrac{14}{3}}$

Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $3x^{2}+19x-154=0$.Therefore, $a=3$,$b=19$ and $c=-154$ Substitute the values: $x=\dfrac{-19±{\sqrt {19^{2}-4(3)(-154)}}}{2(3)}$ =$\dfrac{-19±{\sqrt {361+1848}}}{6}$ =$\dfrac{-19± {\sqrt {2209}}}{6}$ =$\dfrac{-19±47}{6}$ Separate the equation into plus and minus equations: $x=\dfrac{-19-17}{6}=-11$ $x=\dfrac{-19+47}{6}={\dfrac{14}{3}}$
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