jmc

First, we note that $x$ must be nonnegative, since $\sqrt{12x}$ is undefined if $x<0$. Then, we simplify both sides of the equation. Expanding the product on the left gives $$\begin{array}{rl}(\sqrt{12x}+12)(\sqrt{3x}-6)& =\sqrt{12x}(\sqrt{3x}-6)+12(\sqrt{3x}-6)\\ & =\sqrt{36x^2}-6\sqrt{12x}+12\sqrt{3x}-72.\end{array}$$Next, we notice that since $x>0$, we have $\sqrt{36x^2}=6x$. Also, we have $\sqrt{12x}=\sqrt{4\cdot 3x}=2\sqrt{3x}$, so $$\sqrt{36x^2}-6\sqrt{12x}+12\sqrt{3x}-72=6x-6(2\sqrt{3x})+12\sqrt{3x}-72=6x-72.$$Therefore, the left side of the original equation is equivalent to $6x-72$. Simplifying the right side then gives $$6x-72=5x-22.$$Then we collect like terms, to get: $$x=\boxed{50}.$$