Solve sqrt{60^2+(60sqrt{3})^2-628}

$$\sqrt { 60 ^ { 2 } + ( 60 \sqrt { 3 } ) ^ { 2 } - 628 }$$

Evaluate

$$2\sqrt{3443}\approx 117.354164817$$

Solution Steps

Calculate 60 to the power of 2 and get 3600.

$$\sqrt{3600+\left(60\sqrt{3}\right)^{2}-628}$$

Expand \left(60\sqrt{3}\right)^{2}.

$$\sqrt{3600+60^{2}\left(\sqrt{3}\right)^{2}-628}$$

Calculate 60 to the power of 2 and get 3600.

$$\sqrt{3600+3600\left(\sqrt{3}\right)^{2}-628}$$

The square of \sqrt{3} is 3.

$$\sqrt{3600+3600\times 3-628}$$

Multiply 3600 and 3 to get 10800.

$$\sqrt{3600+10800-628}$$

Add 3600 and 10800 to get 14400.

$$\sqrt{14400-628}$$

Subtract 628 from 14400 to get 13772.

$$\sqrt{13772}$$

Factor 13772=2^{2}\times 3443. Rewrite the square root of the product \sqrt{2^{2}\times 3443} as the product of square roots \sqrt{2^{2}}\sqrt{3443}. Take the square root of 2^{2}.

$$2\sqrt{3443}$$