Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 98


$x=\frac{-3\pm\sqrt {17}}{4}$

Work Step by Step

$2x^2 + 3x = 1$ $2x^2 + 3x -1=0$ $x=\frac{-3\pm\sqrt {3^2-4*2*(-1)}}{2*2}=\frac{-3\pm\sqrt {9+8}}{4}=\frac{-3\pm\sqrt {17}}{4}$
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