Solve {left(-2+sqrt{6}right)}^2+4(-2+sqrt{6})-2

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

Evaluate

$$0$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

Add 4 and 6 to get 10.

$$10-4\sqrt{6}+4\left(-2+\sqrt{6}\right)-2$$

Use the distributive property to multiply 4 by -2+\sqrt{6}.

$$10-4\sqrt{6}-8+4\sqrt{6}-2$$

Subtract 8 from 10 to get 2.

$$2-4\sqrt{6}+4\sqrt{6}-2$$

Combine -4\sqrt{6} and 4\sqrt{6} to get 0.

$$2-2$$

Subtract 2 from 2 to get 0.

$$0$$

Factor

$$0$$