jmc

In order to get rid of the cube root, we start by cubing the both sides of the equation $$\begin{array}{rl}(\sqrt[3]{4x^2}{)}^3& =(4{)}^3\\ \Rightarrow 4x^2& =64\\ \Rightarrow x^2& =16\end{array}$$From here, we can see that the only possible values of $x$ are 4 and -4. Since the question asked to list them from least to greatest, the final answer is $\boxed{-4,4}$.