College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 68


$x=-0.53$ or $x=-3.37$

Work Step by Step

We are given: $x^{2}+3.9x+1.8=0$ We solve using the quadratic formula ($a=1,\ b=3.9,\ c=1.8$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\displaystyle \frac{-3.9\pm\sqrt{(3.9)^{2}-4*1*1.8}}{2*1}$ $x=\displaystyle \frac{-3.9\pm\sqrt{15.21-7.2}}{2}$ $x=\displaystyle \frac{-3.9\pm\sqrt{8.01}}{2}$ $x=-0.53$ or $x=-3.37$
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