Answer
$(\sqrt{3}-\sqrt{-4})(\sqrt{6}-\sqrt{-8})=-\sqrt{2}-4\sqrt{6}i$
Work Step by Step
$(\sqrt{3}-\sqrt{-4})(\sqrt{6}-\sqrt{-8})$
Rewrite $\sqrt{-4}$ as $(\sqrt{-1})(\sqrt{4})$ and $\sqrt{-8}$ as $(\sqrt{-1})(\sqrt{8})$:
$[\sqrt{3}-(\sqrt{-1})(\sqrt{4})][\sqrt{6}-(\sqrt{-1})(\sqrt{8})]=...$
Evaluate $\sqrt{4}$ and $\sqrt{8}$:
$...=[\sqrt{3}-(\sqrt{-1})(2)][\sqrt{6}-(\sqrt{-1})(2\sqrt{2})]=...$
Since $\sqrt{-1}=i$, this expression becomes:
$...=(\sqrt{3}-2i)(\sqrt{6}-2\sqrt{2}i)=...$
Evaluate the product and simplify:
$...=(\sqrt{3})(\sqrt{6})-2(\sqrt{3})(\sqrt{2})i-(2)(\sqrt{6})i+(2i)(2\sqrt{2}i)=...$
$...=\sqrt{18}-2\sqrt{6}i-2\sqrt{6}i+4\sqrt{2}(i^{2})=...$
$...=3\sqrt{2}-4\sqrt{6}i+4\sqrt{2}(-1)=...$
$...=3\sqrt{2}-4\sqrt{2}-4\sqrt{6}i=...$
$...=-\sqrt{2}-4\sqrt{6}i$
