Answer
{$\frac{-3 - i\sqrt {39}}{4},\frac{-3 + i\sqrt {39}}{4}$}
Work Step by Step
Step 1: Comparing $-2x^{2}-3x-6=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=-2$, $b=-3$ and $c=-6$
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:
$x=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(-2)(-6)}}{2(-2)}$
Step 4: $x=\frac{3 \pm \sqrt {9-48}}{-4}$
Step 5: $x=\frac{3 \pm \sqrt {-39}}{-4}$
Step 6: $x=\frac{3 \pm \sqrt {-1\times39}}{-4}$
Step 7: $x=\frac{3 \pm (\sqrt {-1}\times\sqrt {39})}{-4}$
Step 8: $x=\frac{3 \pm (i\times \sqrt {39})}{-4}$
Step 9: $x=\frac{3 \pm i\sqrt {39}}{-4}$
Step 10: $x=\frac{3 - i\sqrt {39}}{-4}$ or $x=\frac{3 + i\sqrt {39}}{-4}$
Step 11: $x=\frac{-(3 - i\sqrt {39})}{4}$ or $x=\frac{-(3 + i\sqrt {39})}{4}$
Step 12: Therefore, the solution set is {$\frac{-3 - i\sqrt {39}}{4},\frac{-3 + i\sqrt {39}}{4}$}.
