Solve (sqrt{2}-3)^2+6*(sqrt{2}-3)+5

$$( \sqrt { 2 } - 3 ) ^ { 2 } + 6 \times ( \sqrt { 2 } - 3 ) + 5$$

Evaluate

$$-2$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-6\sqrt{2}+9+6\left(\sqrt{2}-3\right)+5$$

The square of \sqrt{2} is 2.

$$2-6\sqrt{2}+9+6\left(\sqrt{2}-3\right)+5$$

Add 2 and 9 to get 11.

$$11-6\sqrt{2}+6\left(\sqrt{2}-3\right)+5$$

Use the distributive property to multiply 6 by \sqrt{2}-3.

$$11-6\sqrt{2}+6\sqrt{2}-18+5$$

Combine -6\sqrt{2} and 6\sqrt{2} to get 0.

$$11-18+5$$

Subtract 18 from 11 to get -7.

$$-7+5$$

Add -7 and 5 to get -2.

$$-2$$

Factor

$$-2$$