Solve (2sqrt{2}-2p)^2-8=0

$$( 2 \sqrt { 2 } - 2 p ) ^ { 2 } - 8 = 0$$

Solve for p

$$p=0$$

$$p=2\sqrt{2}\approx 2.828427125$$

Steps Using Factoring

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

$$4\left(\sqrt{2}\right)^{2}-8\sqrt{2}p+4p^{2}-8=0$$

The square of \sqrt{2} is 2.

$$4\times 2-8\sqrt{2}p+4p^{2}-8=0$$

Multiply 4 and 2 to get 8.

$$8-8\sqrt{2}p+4p^{2}-8=0$$

Subtract 8 from 8 to get 0.

$$-8\sqrt{2}p+4p^{2}=0$$

Factor out p.

$$p\left(-8\sqrt{2}+4p\right)=0$$

To find equation solutions, solve p=0 and -8\sqrt{2}+4p=0.

$$p=0 p=2\sqrt{2}$$