Solve 20+(2-x)^2+4=(x+2)^2

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

Solve for x

$$x=3$$

Steps for Solving Linear Equation

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

$$20+4-4x+x^{2}+4=\left(x+2\right)^{2}$$

Add 20 and 4 to get 24.

$$24-4x+x^{2}+4=\left(x+2\right)^{2}$$

Add 24 and 4 to get 28.

$$28-4x+x^{2}=\left(x+2\right)^{2}$$

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

$$28-4x+x^{2}=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$28-4x+x^{2}-x^{2}=4x+4$$

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

$$28-4x=4x+4$$

Subtract 4x from both sides.

$$28-4x-4x=4$$

Combine -4x and -4x to get -8x.

$$28-8x=4$$

Subtract 28 from both sides.

$$-8x=4-28$$

Subtract 28 from 4 to get -24.

$$-8x=-24$$

Divide both sides by -8.

$$x=\frac{-24}{-8}$$

Divide -24 by -8 to get 3.

$$x=3$$