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: 22

Answer

{$-3 - \sqrt {14},-3 + \sqrt {14}$}

Work Step by Step

Step 1: Comparing $t^{2}+6t-5=0$ to the standard form of a quadratic equation $at^{2}+bt+c=0$, we obtain: $a=1$, $b=6$ and $c=-5$ 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: $x=\frac{-(6) \pm \sqrt {(6)^{2}-4(1)(-5)}}{2(1)}$ Step 4: $x=\frac{-6 \pm \sqrt {36+20}}{2}$ Step 5: $x=\frac{-6 \pm \sqrt {56}}{2}$ Step 6: $x=\frac{-6 \pm \sqrt {4\times14}}{2}$ Step 7: $x=\frac{-6 \pm \sqrt {2^{2}\times14}}{2}$ Step 8: $x=\frac{-6 \pm 2\sqrt {14}}{2}$ Step 9: $x=-3 \pm \sqrt {14}$ Step 10: $x=-3 + \sqrt {14}$ or $x=-3 - \sqrt {14}$ Step 11: Therefore, the solution set is {$-3 - \sqrt {14},-3 + \sqrt {14}$}.
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