Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4 - Page 457: 15



Work Step by Step

Factoring the equation and setting the factors equal to zero, we obtain: $5x^{2}-2x-7=0$ $5x^{2}+5x-7x-7=0$ $5x(x+1)-7(x+1)=0$ $(x+1)(5x-7)=0$ $(x+1)=0$ 0r $(5x-7)=0$ $x=-1$ or $x=\frac{7}{5}$ Therefore, the solution set is {$-1,\frac{7}{5}$}.
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