Answer
\begin{align*}
x&=-1 &\text{or}& &x=2
\end{align*}
Work Step by Step
Add $1$ to both sides:
\begin{align*}
\sqrt{3x+3}&=x+1
\end{align*}
Square both sides.
Note that $(x+a)^2=x^2+2ax+a^2$.
\begin{align*}
\left(\sqrt{3x+3}\right)^2&=(x+1)^2\\
3x+3&=x^2+2x+1
\end{align*}
Put all terms on the right side:
\begin{align*}
0&=x^2+2x+1-(3x+3)\\
0&=x^2+2x+1-3x-3\\
0&=x^2-x-2\\
\end{align*}
Factor the trinomial to obtain:
\begin{align*}
0&=(x-2)(x+1)
\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x-2&=0 &\text{or}& &x+1=0\\
x&=2 & \text{or}& &x=-1
\end{align*}