Solve P=(sqrt{3}+2)^2+(sqrt{3}-1)^2-2sqrt{3}

$$P = ( \sqrt { 3 } + 2 ) ^ { 2 } + ( \sqrt { 3 } - 1 ) ^ { 2 } - 2 \sqrt { 3 }$$

Solve for P

$$P=11$$

Steps for Solving Linear Equation

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

$$P=\left(\sqrt{3}\right)^{2}+4\sqrt{3}+4+\left(\sqrt{3}-1\right)^{2}-2\sqrt{3}$$

The square of \sqrt{3} is 3.

$$P=3+4\sqrt{3}+4+\left(\sqrt{3}-1\right)^{2}-2\sqrt{3}$$

Add 3 and 4 to get 7.

$$P=7+4\sqrt{3}+\left(\sqrt{3}-1\right)^{2}-2\sqrt{3}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$P=7+4\sqrt{3}+\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1-2\sqrt{3}$$

The square of \sqrt{3} is 3.

$$P=7+4\sqrt{3}+3-2\sqrt{3}+1-2\sqrt{3}$$

Add 3 and 1 to get 4.

$$P=7+4\sqrt{3}+4-2\sqrt{3}-2\sqrt{3}$$

Add 7 and 4 to get 11.

$$P=11+4\sqrt{3}-2\sqrt{3}-2\sqrt{3}$$

Combine 4\sqrt{3} and -2\sqrt{3} to get 2\sqrt{3}.

$$P=11+2\sqrt{3}-2\sqrt{3}$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$P=11$$

Assign P

$$P≔11$$