Solve 4+x^2=left(x+2right)^2

$$4+ { x }^{ 2 } = { \left(x+2 \right) }^{ 2 }$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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

$$4+x^{2}=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$4+x^{2}-x^{2}=4x+4$$

Combine x^{2} and -x^{2} to get 0.

$$4=4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$4x+4=4$$

Subtract 4 from both sides.

$$4x=4-4$$

Subtract 4 from 4 to get 0.

$$4x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 4 is not equal to 0, x must be equal to 0.

$$x=0$$