Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Appendix A - Equations and Inequalities - Page 417: 63

Answer

The solution is {$\frac{-2 \pm \sqrt{10}}{2}$}

Work Step by Step

$0.2x^2$ $+ 0.4x$ $- 0.3$ = $0$ $2x^2$ $+ 4x$ $- 3$ = $0$ $x$ $=$ $\frac{-4 \pm \sqrt{4^2 - (4)(2)(-3)}}{2 (2)}$ (Quadratic formula) $x$ $=$ $\frac{-4 \pm \sqrt{16 + 24}}{4}$ $x$ $=$ $\frac{-4 \pm 2\sqrt{10}}{4}$ $x$ $=$ $\frac{-2 \pm \sqrt{10}}{2}$ The solution is {$\frac{-2 \pm \sqrt{10}}{2}$}
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