Algebra 1: Common Core (15th Edition)

The answer is $12-8\sqrt3$.
We can simplify each product as follows: $=-\sqrt12(4-2\sqrt3)$ $=-\sqrt12*4+2*\sqrt3*\sqrt12$ $=-\sqrt4*\sqrt3*4+2*\sqrt3*\sqrt4*\sqrt3$ $=-2*\sqrt3*4+2*\sqrt3*2*\sqrt3$ $=-8*\sqrt3+4*3$ $=-8\sqrt3+12$ $=12-8\sqrt3$.