(2+x)^2-(2x-4)^2+4x-2=2^2+x^2+2*2*x

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Solution for (2+x)^2-(2x-4)^2+4x-2=2^2+x^2+2*2*x equation: 



(2+x)^2-(2x-4)^2+4x-2=2^2+x^2+2*2x

We move all terms to the left:

(2+x)^2-(2x-4)^2+4x-2-(2^2+x^2+2*2x)=0

We add all the numbers together, and all the variables

-(2^2+x^2+2*2x)+(x+2)^2-(2x-4)^2+4x-2=0

We add all the numbers together, and all the variables

-(2^2+x^2+2*2x)+4x+(x+2)^2-(2x-4)^2-2=0

We get rid of parentheses

-x^2-2*2x+4x+(x+2)^2-(2x-4)^2-2-2^2=0

We add all the numbers together, and all the variables

-1x^2+4x-2*2x+(x+2)^2-(2x-4)^2-6=0

Wy multiply elements

-1x^2+4x-4x+(x+2)^2-(2x-4)^2-6=0

We add all the numbers together, and all the variables

-1x^2+(x+2)^2-(2x-4)^2-6=0

We move all terms containing x to the left, all other terms to the right

-1x^2+(x+2)^2-(2x-4)^2=6

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