(x+1)(x+2)-3=(x-3)(x-5)

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Solution for (x+1)(x+2)-3=(x-3)(x-5) equation: 



(x+1)(x+2)-3=(x-3)(x-5)

We move all terms to the left:

(x+1)(x+2)-3-((x-3)(x-5))=0

We multiply parentheses ..

(+x^2+2x+x+2)-((x-3)(x-5))-3=0



We calculate terms in parentheses: -((x-3)(x-5)), so:

(x-3)(x-5)

We multiply parentheses ..

(+x^2-5x-3x+15)

We get rid of parentheses

x^2-5x-3x+15

We add all the numbers together, and all the variables

x^2-8x+15

Back to the equation:

-(x^2-8x+15)



We get rid of parentheses

x^2-x^2+2x+x+8x+2-15-3=0

We add all the numbers together, and all the variables

11x-16=0

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

11x=16

x=16/11

x=1+5/11

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