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: 32

Answer

{$\frac{-1-\sqrt {13}}{3},\frac{-1+\sqrt {13}}{3}$}

Work Step by Step

Step 1: Comparing $3x^{2}+2x-4=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=3$, $b=2$ and $c=-4$ 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: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(3)(-4)}}{2(3)}$ Step 4: $x=\frac{-2 \pm \sqrt {4+48}}{6}$ Step 5: $x=\frac{-2 \pm \sqrt {52}}{6}$ Step 6: $x=\frac{-2 \pm \sqrt {4\times13}}{6}$ Step 7: $x=\frac{-2 \pm 2\sqrt {13}}{6}$ Step 8: $x=\frac{-1 \pm \sqrt {13}}{3}$ Step 9: $x=\frac{-1-\sqrt {13}}{3}$ or $x=\frac{-1+\sqrt {13}}{3}$ Step 10: Therefore, the solution set is {$\frac{-1-\sqrt {13}}{3},\frac{-1+\sqrt {13}}{3}$}.
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