Answer
$3x-2\sqrt{6x}+2$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$, or the square of a binomial, and the properties of radicals, the given expression, $
(\sqrt{3x}-\sqrt{2})^2
,$ is equivalent to
\begin{array}{l}\require{cancel}
(\sqrt{3x})^2+2(\sqrt{3x})(-\sqrt{2})+(-\sqrt{2})^2
\\\\=
3x-2\sqrt{6x}+2
.\end{array}