Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.5 - Quadratic Equations: Complex Solutions - Problem Set 11.5 - Page 497: 21



Work Step by Step

Using the rules of factoring trinomials to factor, we obtain: $2x^{2}-3x-5=0$ $2x^{2}+2x-5x-5=0$ $2x(x+1)-5(x+1)=0$ $(x+1)(2x-5)=0$ $(x+1)=0$ or $(2x-5)=0$ $x=-1$ or $x=\frac{5}{2}$ Therefore, the solution set is {$-1,\frac{5}{2}$}.
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