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 458: 26



Work Step by Step

Factoring the equation and setting the factors equal to zero, we obtain: $2x^{2}+10x-28=0$ $2x^{2}-4x+14x-28=0$ $2x(x-2)+14(x-2)=0$ $(x-2)(2x+14)=0$ $(x-2)=0$ or $(2x+14)=0$ $x=2$ or $x=-\frac{14}{2}$ $x=2$ or $x=-7$ Therefore, the solution set is {$-7,2$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.