Solve left(x+2right)^2=x^2+20

$${ \left(x+2 \right) }^{ 2 } = { x }^{ 2 } +20$$

Solve for x

$$x=4$$

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}.

$$x^{2}+4x+4=x^{2}+20$$

Subtract x^{2} from both sides.

$$x^{2}+4x+4-x^{2}=20$$

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

$$4x+4=20$$

Subtract 4 from both sides.

$$4x=20-4$$

Subtract 4 from 20 to get 16.

$$4x=16$$

Divide both sides by 4.

$$x=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$x=4$$