Answer
{$-3 - i\sqrt 3,-3 + i\sqrt 3$}
Work Step by Step
Step 1: Comparing $t^{2}+6t+12=0$ to the standard form of a quadratic equation, $at^{2}+bt+c=0$, we find:
$a=1$, $b=6$ and $c=12$
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:
$x=\frac{-(6) \pm \sqrt {(6)^{2}-4(1)(12)}}{2(1)}$
Step 4: $x=\frac{-6 \pm \sqrt {36-48}}{2}$
Step 5: $x=\frac{-6 \pm \sqrt {-12}}{2}$
Step 6: $x=\frac{-6 \pm \sqrt {-1\times12}}{2}$
Step 7: $x=\frac{-6 \pm (\sqrt {-1}\times\sqrt {4\times3})}{2}$
Step 8: $x=\frac{-6 \pm (i\times 2\sqrt 3)}{2}$
Step 9: $x=\frac{-6 \pm i2\sqrt 3}{2}$
Step 10: $x=\frac{2(-3 \pm i\sqrt 3)}{2}$
Step 11: $x=-3 \pm i\sqrt 3$
Step 12: $x=-3 - i\sqrt 3$ or $x=-3 + i\sqrt 3$
Step 13: Therefore, the solution set is {$-3 - i\sqrt 3,-3 + i\sqrt 3$}.
