Evaluate
\sqrt{3}+9-4\sqrt{2}\approx 5.075196558
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\sqrt{3}+4\left(\sqrt{2}\right)^{2}-4\sqrt{2}+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-1\right)^{2}.
\sqrt{3}+4\times 2-4\sqrt{2}+1
The square of \sqrt{2} is 2.
\sqrt{3}+8-4\sqrt{2}+1
Multiply 4 and 2 to get 8.
\sqrt{3}+9-4\sqrt{2}
Add 8 and 1 to get 9.
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