Solve p=(3-sqrt{2})^2+(2sqrt{2}-3)^2

$$p = ( 3 - \sqrt { 2 } ) ^ { 2 } + ( 2 \sqrt { 2 } - 3 ) ^ { 2 }$$

Solve for p

$$p=28-18\sqrt{2}\approx 2.544155877$$

Steps for Solving Linear Equation

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

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

The square of \sqrt{2} is 2.

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

Add 9 and 2 to get 11.

$$p=11-6\sqrt{2}+\left(2\sqrt{2}-3\right)^{2}$$

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

$$p=11-6\sqrt{2}+4\left(\sqrt{2}\right)^{2}-12\sqrt{2}+9$$

The square of \sqrt{2} is 2.

$$p=11-6\sqrt{2}+4\times 2-12\sqrt{2}+9$$

Multiply 4 and 2 to get 8.

$$p=11-6\sqrt{2}+8-12\sqrt{2}+9$$

Add 8 and 9 to get 17.

$$p=11-6\sqrt{2}+17-12\sqrt{2}$$

Add 11 and 17 to get 28.

$$p=28-6\sqrt{2}-12\sqrt{2}$$

Combine -6\sqrt{2} and -12\sqrt{2} to get -18\sqrt{2}.

$$p=28-18\sqrt{2}$$

Assign p

$$p≔28-18\sqrt{2}$$