Solve (20-sqrt{12})^2

$$( 20 - \sqrt { 12 } ) ^ { 2 }$$

Evaluate

$$412-80\sqrt{3}\approx 273.435935394$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(20-2\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(20-2\sqrt{3}\right)^{2}.

$$400-80\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$400-80\sqrt{3}+4\times 3$$

Multiply 4 and 3 to get 12.

$$400-80\sqrt{3}+12$$

Add 400 and 12 to get 412.

$$412-80\sqrt{3}$$

Expand

$$412-80\sqrt{3}$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(20-2\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(20-2\sqrt{3}\right)^{2}.

$$400-80\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$400-80\sqrt{3}+4\times 3$$

Multiply 4 and 3 to get 12.

$$400-80\sqrt{3}+12$$

Add 400 and 12 to get 412.

$$412-80\sqrt{3}$$