Solve {left(3sqrt{6}-2right)}^2+12sqrt{6}

$${ \left(3 \sqrt{ 6 } -2 \right) }^{ 2 } +12 \sqrt{ 6 }$$

Evaluate

$$58$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{6}-2\right)^{2}.

$$9\left(\sqrt{6}\right)^{2}-12\sqrt{6}+4+12\sqrt{6}$$

The square of \sqrt{6} is 6.

$$9\times 6-12\sqrt{6}+4+12\sqrt{6}$$

Multiply 9 and 6 to get 54.

$$54-12\sqrt{6}+4+12\sqrt{6}$$

Add 54 and 4 to get 58.

$$58-12\sqrt{6}+12\sqrt{6}$$

Combine -12\sqrt{6} and 12\sqrt{6} to get 0.

$$58$$

Factor

$$2\times 29$$