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