Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Cumulative Review - Page 508: 1

Answer

$ \frac{-1\pm\sqrt {13}}{6}$

Work Step by Step

Using the quadratic formula: $3x^2+x-1=0 \longrightarrow x=\frac{-1\pm\sqrt {1+4(3)}}{2(3)}=\frac{-1\pm\sqrt {13}}{6}$
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