Solve (5-2sqrt{6}*5+2sqrt{6})^2

$$( 5 - 2 \sqrt { 6 } \cdot 5 + 2 \sqrt { 6 } ) ^ { 2 }$$

Evaluate

$$409-80\sqrt{6}\approx 213.040820577$$

Solution Steps

Multiply 2 and 5 to get 10.

$$\left(5-10\sqrt{6}+2\sqrt{6}\right)^{2}$$

Combine -10\sqrt{6} and 2\sqrt{6} to get -8\sqrt{6}.

$$\left(5-8\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-8\sqrt{6}\right)^{2}.

$$25-80\sqrt{6}+64\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$25-80\sqrt{6}+64\times 6$$

Multiply 64 and 6 to get 384.

$$25-80\sqrt{6}+384$$

Add 25 and 384 to get 409.

$$409-80\sqrt{6}$$

Expand

$$409-80\sqrt{6}$$

Solution Steps

Multiply 2 and 5 to get 10.

$$\left(5-10\sqrt{6}+2\sqrt{6}\right)^{2}$$

Combine -10\sqrt{6} and 2\sqrt{6} to get -8\sqrt{6}.

$$\left(5-8\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-8\sqrt{6}\right)^{2}.

$$25-80\sqrt{6}+64\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$25-80\sqrt{6}+64\times 6$$

Multiply 64 and 6 to get 384.

$$25-80\sqrt{6}+384$$

Add 25 and 384 to get 409.

$$409-80\sqrt{6}$$