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: 60

Answer

{$-1.61,2.28$}

Work Step by Step

Step 1: Comparing $-3x^{2}+2x+11=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=-3$, $b=2$ and $c=11$ 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: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(-3)(11)}}{2(-3)}$ Step 4: $x=\frac{-2 \pm \sqrt {4+132}}{-6}$ Step 5: $x=\frac{-2 \pm \sqrt {136}}{-6}$ Step 6: $x=\frac{-2 \pm 11.662}{-6}$ Step 7: $x=\frac{-2 + 11.662}{-6}$ or $x=\frac{-2 - 11.662}{-6}$ Step 8: $x=\frac{9.662}{-6}$ or $x=\frac{-13.662}{-6}$ Step 9: $x=-1.610\approx-1.61$ or $x=2.277\approx2.28$ Step 10: Therefore, the solution set is {$-1.61,2.28$}.
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