Answer
$\color{blue}{x = \left\{-2, 0\right\}}$
Work Step by Step
Let $u = x+3$
Replacing $x+3$ with $u$ gves:
$(x+3)^2-4(x+3)+3=0
\\u^2-4u+3=0$
Factor the trinomial to obtain:
$(u-3)(u-1)=0$
Use the Zero-Product Property (which states that if $xy=0$, then either $x=0$ or $y=0$ or both are zero) by equating each factor to zero to obtain:
$u-3=0 \text{ or } u-1=0$
Solve each equation to obtain:
$u=3$ or $u=1$
Replace $u$ with $x+3$ to obtain:
\begin{array}{ccc}
&u=3 &\text{or} &u=1
\\&x+3=3 &\text{or} &x+3=1
\\&x=3-3 &\text{or} &x=1-3
\\&x=0 &\text{or} &x=-2
\end{array}
Therefore, the solutions to the given equation are:
$\color{blue}{x = \left\{-2, 0\right\}}$
