Evaluate
14\sqrt{3}+52\approx 76.248711306
Expand
14 \sqrt{3} + 52 = 76.248711306
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\left(\sqrt{3}\right)^{2}+14\sqrt{3}+49
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+7\right)^{2}.
3+14\sqrt{3}+49
The square of \sqrt{3} is 3.
52+14\sqrt{3}
Add 3 and 49 to get 52.
\left(\sqrt{3}\right)^{2}+14\sqrt{3}+49
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+7\right)^{2}.
3+14\sqrt{3}+49
The square of \sqrt{3} is 3.
52+14\sqrt{3}
Add 3 and 49 to get 52.
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