Solve x^2=(x+2)^2+16

$$x ^ { 2 } = ( x + 2 ) ^ { 2 } + 16$$

Solve for x

$$x=-5$$

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}.

$$x^{2}=x^{2}+4x+4+16$$

Add 4 and 16 to get 20.

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

Subtract x^{2} from both sides.

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

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

$$0=4x+20$$

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

$$4x+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$4x=-20$$

Divide both sides by 4.

$$x=\frac{-20}{4}$$

Divide -20 by 4 to get -5.

$$x=-5$$