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

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



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

We move all terms to the left:

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

We multiply parentheses ..

(+x^2+2x-3x-6)(x-5)-((+x^2+2x-3x-6))=0



We calculate terms in parentheses: -((+x^2+2x-3x-6)), so:

(+x^2+2x-3x-6)

We get rid of parentheses

x^2+2x-3x-6

We add all the numbers together, and all the variables

x^2-1x-6

Back to the equation:

-(x^2-1x-6)



We get rid of parentheses

(+x^2+2x-3x-6)(x-5)-x^2+1x+6=0

We add all the numbers together, and all the variables

-1x^2+(+x^2+2x-3x-6)(x-5)+x+6=0

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

-1x^2+(+x^2+2x-3x-6)(x-5)+x=-6

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