Answer
{$\frac{-1 + i\sqrt {35}}{6},\frac{-1 - i\sqrt {35}}{6}$}
Work Step by Step
$p=-3p^{2}-3$ can be rearranged as $3p^{2}+p+3=0$
Now solving $3p^{2}+p+3=0$:
Step 1: Comparing $3p^{2}+p+3=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=3$, $b=1$ 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:
$x=\frac{-(1) \pm \sqrt {1^{2}-4(3)(3)}}{2(3)}$
Step 4: $x=\frac{-1 \pm \sqrt {1-36}}{6}$
Step 5: $x=\frac{-1 \pm \sqrt {-35}}{6}$
Step 6: $x=\frac{-1 \pm \sqrt {-1\times35}}{6}$
Step 7: $x=\frac{-1 \pm \sqrt {-1}\times\sqrt {35}}{6}$
Step 8: $x=\frac{-1 \pm i\sqrt {35}}{6}$
Step 9: $x=\frac{-1 + i\sqrt {35}}{6}$ or $x=\frac{-1 - i\sqrt {35}}{6}$
Step 10: Therefore, the solution set is {$\frac{-1 + i\sqrt {35}}{6},\frac{-1 - i\sqrt {35}}{6}$}.
