If $\sqrt{3x-5}=2$, find all possible values of $x$.
First, we start by squaring both sides of the equation \begin{align*} (\sqrt{3x-5})^2& =(2)^2
\\ \Rightarrow\qquad 3x-5& =4
\\\Rightarrow\qquad 3x& =9
\\\Rightarrow\qquad x& =\boxed{3}.
\end{align*}Testing, we find that this value of $x$ does indeed satisfy the equation.