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7
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7
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\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4+4\sqrt{3}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-2\right)^{2}.
3-4\sqrt{3}+4+4\sqrt{3}
The square of \sqrt{3} is 3.
7-4\sqrt{3}+4\sqrt{3}
Add 3 and 4 to get 7.
7
Combine -4\sqrt{3} and 4\sqrt{3} to get 0.
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