Answer
{$\frac{-3 - \sqrt {33}}{4}\frac{-3 + \sqrt {33}}{4}$}
Work Step by Step
Step 1: Comparing $2x^{2}+3x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=2$, $b=3$ and $c=-3$
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:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(-3)}}{2(2)}$
Step 4: $x=\frac{-3 \pm \sqrt {9+24}}{4}$
Step 5: $x=\frac{-3 \pm \sqrt {33}}{4}$
Step 6: $x=\frac{-3 - \sqrt {33}}{4}$ or $x=\frac{-3 + \sqrt {33}}{4}$
Step 7: Therefore, the solution set is {$\frac{-3 - \sqrt {33}}{4},\frac{-3 + \sqrt {33}}{4}$}.
