## Elementary Algebra

Published by Cengage Learning

# Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3 - Page 453: 31

#### Answer

{$\frac{-2-\sqrt {7}}{3},\frac{-2+\sqrt {7}}{3}$}

#### Work Step by Step

Step 1: Comparing $3x^{2}+4x-1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=3$, $b=4$ and $c=-1$ 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: $x=\frac{-(4) \pm \sqrt {(4)^{2}-4(3)(-1)}}{2(3)}$ Step 4: $x=\frac{-4 \pm \sqrt {16+12}}{6}$ Step 5: $x=\frac{-4 \pm \sqrt {28}}{6}$ Step 6: $x=\frac{-4 \pm \sqrt {4\times7}}{6}$ Step 7: $x=\frac{-4 \pm 2\sqrt {7}}{6}$ Step 8: $x=\frac{-2 \pm \sqrt {7}}{3}$ Step 9: $x=\frac{-2-\sqrt {7}}{3}$ or $x=\frac{-2+\sqrt {7}}{3}$ Step 10: Therefore, the solution set is {$\frac{-2-\sqrt {7}}{3},\frac{-2+\sqrt {7}}{3}$}.

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