(x-3)(2x-5)+4(2-x)+12=(2-2x)(1-x)

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Solution for (x-3)(2x-5)+4(2-x)+12=(2-2x)(1-x) equation: 



(x-3)(2x-5)+4(2-x)+12=(2-2x)(1-x)

We move all terms to the left:

(x-3)(2x-5)+4(2-x)+12-((2-2x)(1-x))=0

We add all the numbers together, and all the variables

(x-3)(2x-5)+4(-1x+2)-((-2x+2)(-1x+1))+12=0

We multiply parentheses

(x-3)(2x-5)-4x-((-2x+2)(-1x+1))+8+12=0

We multiply parentheses ..

(+2x^2-5x-6x+15)-4x-((-2x+2)(-1x+1))+8+12=0



We calculate terms in parentheses: -((-2x+2)(-1x+1)), so:

(-2x+2)(-1x+1)

We multiply parentheses ..

(+2x^2-2x-2x+2)

We get rid of parentheses

2x^2-2x-2x+2

We add all the numbers together, and all the variables

2x^2-4x+2

Back to the equation:

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



We add all the numbers together, and all the variables

(+2x^2-5x-6x+15)-4x-(2x^2-4x+2)+20=0

We get rid of parentheses

2x^2-2x^2-5x-6x-4x+4x+15-2+20=0

We add all the numbers together, and all the variables

-11x+33=0

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

-11x=-33

x=-33/-11

x=+3

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