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: 35



Work Step by Step

Factoring the equation and setting the factors equal to zero, we obtain: $t^{2}+12t+36=49$ $t^{2}+12t+36-49=0$ $t^{2}+12t-13=0$ $t^{2}-1t+13t-13=0$ $t(t-1)+13(t-1)=0$ $(t-1)(t+13)=0$ $(t-1)=0$ or $(t+13)=0$ $t=1$ or $t=-13$ Therefore, the solution set is {$-13,1$}.
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