Step-by-step explanation:

[tex]2 \sqrt{3} - 4 \sqrt{12} - 3 \sqrt{4} [/tex]

1. The sqaureroot of 12 can be broken into the sqaureroot of quantity4*3 and the sqaureroot of 4 is equal to 2

[tex]2 \sqrt{3} - 4 \sqrt{4 \times 3} - 3 \sqrt{4} \\ 2 \sqrt{3} - 4 \times 2 \sqrt{3} - 3 \times 2 \\ 2 \sqrt{3} - 8 \sqrt{3} - 6[/tex]

2. When adding sqaure roots, we can treat them as a variable with base coefficient 1 (1x = x), since the terms 2√3 and -8√3 have the same square roots, we can add them up

[tex]2 \sqrt{3} - 8 \sqrt{3} = 2x - 8x \\ 2x - 8x = - 6x[/tex]

Therefore

[tex]2 \sqrt{3} - 8 \sqrt{3} = - 6 \sqrt{3} [/tex]

Thus the answer is

[tex] - 6 \sqrt{3} - 6[/tex]