Answer
$7\sqrt{6}+8\sqrt{5}$
Work Step by Step
Factor the radicands so that one of the factors is a perfect square:
$=3\sqrt{9(6)}-2\sqrt{4(5)}+4\sqrt{9(5)}-\sqrt{4(6)}
\\=3\sqrt{3^2(6)}-2\sqrt{2^2(5)}+4\sqrt{3^2(5)}-\sqrt{2^2(6)}$
Simplify to obtain:
$\\=3\cdot 3\sqrt{6}-2\cdot 2\sqrt{5}+4\cdot 3\sqrt{5}-2\sqrt{6}
\\=9\sqrt{6}-4\sqrt{5}+12\sqrt{5}-2\sqrt{6}$
Combine like terms:
$=(9\sqrt{6} - 2\sqrt{6}) + (-4\sqrt{5}+12\sqrt{5})
\\=(9-2)\sqrt{6} + (-4+12)\sqrt{5}
\\=7\sqrt{6}+8\sqrt{5}$
