(x+2)(x+2)-(x-2)(x-2)=(x)(x)-(x-5)(x-5)+5

Solution for (x+2)(x+2)-(x-2)(x-2)=(x)(x)-(x-5)(x-5)+5 equation:

(x+2)(x+2)-(x-2)(x-2)=(x)(x)-(x-5)(x-5)+5

We move all terms to the left:

(x+2)(x+2)-(x-2)(x-2)-((x)(x)-(x-5)(x-5)+5)=0

We multiply parentheses ..

(+x^2+2x+2x+4)-(x-2)(x-2)-(xx-(x-5)(x-5)+5)=0


We calculate terms in parentheses: -(xx-(x-5)(x-5)+5), so:

xx-(x-5)(x-5)+5

We multiply parentheses ..

-(+x^2-5x-5x+25)+xx+5

We get rid of parentheses

-x^2+5x+5x+xx-25+5

We add all the numbers together, and all the variables

-1x^2+10x+xx-20

Back to the equation:

-(-1x^2+10x+xx-20)


We get rid of parentheses

x^2+1x^2+2x+2x-10x-xx-(x-2)(x-2)+4+20=0

We multiply parentheses ..

x^2+1x^2-(+x^2-2x-2x+4)+2x+2x-10x-xx+4+20=0

We add all the numbers together, and all the variables

2x^2-(+x^2-2x-2x+4)-6x-xx+24=0

We get rid of parentheses

2x^2-x^2+2x+2x-6x-xx-4+24=0

We add all the numbers together, and all the variables

x^2-2x-xx+20=0

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

x^2-2x-xx=-20