Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 48

Answer

{$-15,13$}

Work Step by Step

Step 1: Comparing $n^{2}+2n-195=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=2$, and $c=-195$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b, and c in the formula, we obtain: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(-195)}}{2(1)}$ Step 4: $x=\frac{-2 \pm \sqrt {4+780}}{2}$ Step 5: $x=\frac{-2 \pm \sqrt {784}}{2}$ Step 6: $x=\frac{-2 \pm 28}{2}$ Step 7: $x=\frac{-2+28}{2}$ or $x=\frac{-2-28}{2}$ Step 8: $x=\frac{26}{2}$ or $x=\frac{-30}{2}$ Step 9: $x=13$ or $x=-15$ Step 10: Therefore, the solution set is {$-15,13$}.
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