Solve {left(3-sqrt{10}right)}^2-6*(3-sqrt{10})-2

$${ \left(3- \sqrt{ 10 } \right) }^{ 2 } -6 \times (3- \sqrt{ 10 } )-2$$

Evaluate

$$-1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{10}\right)^{2}.

$$9-6\sqrt{10}+\left(\sqrt{10}\right)^{2}-6\left(3-\sqrt{10}\right)-2$$

The square of \sqrt{10} is 10.

$$9-6\sqrt{10}+10-6\left(3-\sqrt{10}\right)-2$$

Add 9 and 10 to get 19.

$$19-6\sqrt{10}-6\left(3-\sqrt{10}\right)-2$$

Use the distributive property to multiply -6 by 3-\sqrt{10}.

$$19-6\sqrt{10}-18+6\sqrt{10}-2$$

Subtract 18 from 19 to get 1.

$$1-6\sqrt{10}+6\sqrt{10}-2$$

Combine -6\sqrt{10} and 6\sqrt{10} to get 0.

$$1-2$$

Subtract 2 from 1 to get -1.

$$-1$$

Factor

$$-1$$