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