(x+3)(x+1)-(x+4)(x+2)=8

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Solution for (x+3)(x+1)-(x+4)(x+2)=8 equation: 



(x+3)(x+1)-(x+4)(x+2)=8

We move all terms to the left:

(x+3)(x+1)-(x+4)(x+2)-(8)=0

We multiply parentheses ..

(+x^2+x+3x+3)-(x+4)(x+2)-8=0

We get rid of parentheses

x^2+x+3x-(x+4)(x+2)+3-8=0

We multiply parentheses ..

x^2-(+x^2+2x+4x+8)+x+3x+3-8=0

We add all the numbers together, and all the variables

x^2-(+x^2+2x+4x+8)+4x-5=0

We get rid of parentheses

x^2-x^2-2x-4x+4x-8-5=0

We add all the numbers together, and all the variables

-2x-13=0

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

-2x=13

x=13/-2

x=-6+1/2

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