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 - Page 453: 44


{$\frac{-1-\sqrt {19}}{6},\frac{-1+\sqrt {19}}{6}$}

Work Step by Step

Step 1: Comparing $6x^{2}+2x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=6$, $b=2$ and $c=-3$ 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: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(6)(-3)}}{2(6)}$ Step 4: $x=\frac{-2 \pm \sqrt {4+72}}{12}$ Step 5: $x=\frac{-2 \pm \sqrt {76}}{12}$ Step 6: $x=\frac{-2 \pm \sqrt {4\times19}}{12}$ Step 7: $x=\frac{-2 \pm 2\sqrt {19}}{12}$ Step 8: $x=\frac{-1 \pm \sqrt {19}}{6}$ Step 9: $x=\frac{-1-\sqrt {19}}{6}$ or $x=\frac{-1+\sqrt {19}}{6}$ Step 10: Therefore, the solution set is {$\frac{-1-\sqrt {19}}{6},\frac{-1+\sqrt {19}}{6}$}.
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