(x+2)(x+3)=(x+2)(x-4)+20

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Solution for (x+2)(x+3)=(x+2)(x-4)+20 equation: 



(x+2)(x+3)=(x+2)(x-4)+20

We move all terms to the left:

(x+2)(x+3)-((x+2)(x-4)+20)=0

We multiply parentheses ..

(+x^2+3x+2x+6)-((x+2)(x-4)+20)=0



We calculate terms in parentheses: -((x+2)(x-4)+20), so:

(x+2)(x-4)+20

We multiply parentheses ..

(+x^2-4x+2x-8)+20

We get rid of parentheses

x^2-4x+2x-8+20

We add all the numbers together, and all the variables

x^2-2x+12

Back to the equation:

-(x^2-2x+12)



We get rid of parentheses

x^2-x^2+3x+2x+2x+6-12=0

We add all the numbers together, and all the variables

7x-6=0

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

7x=6

x=6/7

x=6/7

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