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