Answer
$\left\{-\dfrac{8}{3},-1\right\}$
Work Step by Step
Multiplying by the $LCD=
(t+2)^2
,$ the given equation $
3=\dfrac{1}{t+2}+\dfrac{2}{(t+2)^2}
,$ is equivalent to
\begin{align*}\require{cancel}
(t+2)^2(3)&=\left(\dfrac{1}{t+2}+\dfrac{2}{(t+2)^2}\right)(t+2)^2
\\\\
(t^2+4t+4)(3)&=1(t+2)+2(1)
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
3t^2+12t+12&=t+2+2
\\
3t^2+(12t-t)+(12-2-2)&=0
\\
3t^2+11t+8&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(3t+8)(t+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
3t+8=0 & t+1=0
\\
3t=-8 & t=-1
\\\\
t=-\dfrac{8}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }t=-\dfrac{8}{3}: & \text{If }t=-1:
\\\\
3\overset{?}=\dfrac{1}{-\frac{8}{3}+2}+\dfrac{2}{(-\frac{8}{3}+2)^2} &
3\overset{?}=\dfrac{1}{-1+2}+\dfrac{2}{(-1+2)^2}
\\\\
3\overset{?}=\dfrac{1}{-\frac{8}{3}+\frac{6}{3}}+\dfrac{2}{\left(-\frac{8}{3}+\frac{6}{3}\right)^2} &
3\overset{?}=\dfrac{1}{1}+\dfrac{2}{(1)^2}
\\\\
3\overset{?}=\dfrac{1}{-\frac{2}{3}}+\dfrac{2}{\left(-\frac{2}{3}\right)^2} &
3\overset{?}=1+2
\\\\
3\overset{?}=1\cdot\left(-\dfrac{3}{2}\right)+\dfrac{2}{\frac{4}{9}} &
3\overset{\checkmark}=3
\\\\
3\overset{?}=-\dfrac{3}{2}+2\cdot\dfrac{9}{4}
\\\\
3\overset{?}=-\dfrac{3}{2}+\dfrac{9}{2}
\\\\
3\overset{?}=\dfrac{6}{2}
\\\\
3\overset{\checkmark}=3
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
3=\dfrac{1}{t+2}+\dfrac{2}{(t+2)^2}
$ is $\left\{-\dfrac{8}{3},-1\right\}$.
