Answer
{$\frac{2 - \sqrt {10}}{3},\frac{2+\sqrt {10}}{3}$}
Work Step by Step
Step 1: Comparing $3x^{2}-4x-2=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=3$, $b=-4$ and $c=-2$
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:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(3)(-2)}}{2(3)}$
Step 4: $x=\frac{4 \pm \sqrt {16+24}}{6}$
Step 5: $x=\frac{4 \pm \sqrt {40}}{6}$
Step 6: $x=\frac{4 \pm \sqrt {4\times10}}{6}$
Step 7: $x=\frac{4 \pm (\sqrt {4}\times\sqrt {10})}{6}$
Step 8: $x=\frac{4 \pm (2\times \sqrt {10})}{6}$
Step 9: $x=\frac{2(2 \pm 1\sqrt {10})}{6}$
Step 10: $x=\frac{2 - \sqrt {10}}{3}$ or $x=\frac{2 + \sqrt {10}}{3}$
Step 11: Therefore, the solution set is {$\frac{2 - \sqrt {10}}{3},\frac{2+\sqrt {10}}{3}$}.
