[tex]\dfrac{-2\sqrt8}{4+\sqrt{12}}=\dfrac{-2\sqrt{4\cdot2}}{4+\sqrt{4\cdot3}}=\dfrac{-2\cdot\sqrt4\cdot\sqrt2}{4+\sqrt4\cdot\sqrt3}=\dfrac{-2\cdot2\sqrt3}{4+2\sqrt3}=\dfrac{-4\sqrt2}{2(2+\sqrt3)}=-\dfrac{2\sqrt2}{2+\sqrt3}\\\\\text{Use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{-2\sqrt2}{2+\sqrt3}\cdot\dfrac{2-\sqrt3}{2-\sqrt3}=\dfrac{-2\sqrt2(2-\sqrt3)}{2^2-(\sqrt3)^2}\\\\\text{Use distributive property and}\ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\\\\=\dfrac{-4\sqrt2+2\sqrt6}{4-3}=\dfrac{2\sqrt6-4\sqrt2}{1}=2\sqrt6-4\sqrt2[/tex]
