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: 58

Answer

{$-0.90,3.10$}

Work Step by Step

Step 1: Comparing $5x^{2}-11x-14=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=5$, $b=-11$, and $c=-14$ 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: $x=\frac{-(-11) \pm \sqrt {(-11)^{2}-4(5)(-14)}}{2(5)}$ Step 4: $x=\frac{11 \pm \sqrt {121+280}}{10}$ Step 5: $x=\frac{11 \pm \sqrt {401}}{10}$ Step 6: $x=\frac{11 \pm 20.025}{10}$ Step 7: $x=\frac{11+20.025}{10}$ or $x=\frac{11-20.025}{10}$ Step 8: $x=\frac{31.025}{10}$ or $x=\frac{-9.025}{10}$ Step 9: $x=3.1025\approx3.10$ or $x=-0.9025\approx-0.90$ Step 10: Therefore, the solution set is {$-0.90,3.10$}.
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