How to solve (2 + square root of -3)(-1 + square root of -12)

I'm really confused. What I have so far is..

(2 + i square root 3)( -1 + 2i square root 3)

and that's about it. I know that I have to F.O.I.L., but I don't understand how to do it when there are square roots involved.

Square roots multiply in FOIL just like other items do.

First: 2 * -1 = -2

Outside: 2 * 2i sqrt(3) = 4isqrt(3)

Inside: i sqrt(3) * -1 = -i sqrt(3)

Last: i sqrt(3) * 2i sqrt(3) = (i * 2i)[(sqrt(3) * sqrt(3)] = 2i^2 * 3 = 2(-1) * 3 = -6

Combine into:

-2 - 6 + 4i sqrt(3) - i sqrt(3)

Combine like terms:

-8 + 3i sqrt(3) [or perhaps preferably -8 + 3sqrt(3) i]

(2 + i square root 3)( -1 + 2i square root 3) =
-2-isqrt3+4isqrt3+2i^2(sqrt3)^2=
(remember i^2=-1, and (sqrt3)^2=3
-2+3isqrt3-2x3=
-8+3isqrt3

(2+sqrt(-3))(-1+sqrt(-12))
=-2+2sqrt(-12)-sqrt(-3)+sqrt(-3)sqrt(-…
=-2+2i*sqrt(12)-i*sqrt(3)+i*sqrt(3)*i*…
=-2+4i*sqrt(3)-i*sqrt(3)+i^2*6
=-2+3i*sqrt(3)+i^2*6
=-2+3i*sqrt(3)-6
=-8+3*sqrt(3)i

multiply to get -2 +4i.square root(3) -i.square root(3) -2.3

-8 + 3i.square root(3)