Answer
$2-8\sqrt{35}$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$, or the product of 2 binomials, and the properties of radicals, the given expression, $
(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})
,$ is equivalent to
\begin{array}{l}\require{cancel}
(3\sqrt{7})(2\sqrt{7})+(3\sqrt{7})(-4\sqrt{5})+(2\sqrt{5})(2\sqrt{7})+(2\sqrt{5})(-4\sqrt{5})
\\\\=
6\sqrt{49}-12\sqrt{35}+4\sqrt{35}-8\sqrt{25}
\\\\=
6\sqrt{(7)^2}-12\sqrt{35}+4\sqrt{35}-8\sqrt{(5)^2}
\\\\=
6\cdot7-12\sqrt{35}+4\sqrt{35}-8\cdot5
\\\\=
42-12\sqrt{35}+4\sqrt{35}-40
\\\\=
(42-40)+(-12\sqrt{35}+4\sqrt{35})
\\\\=
2-8\sqrt{35}
.\end{array}
