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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-10x^2+25x+4x-10))+(x-3)^3-(x^2-3x+5)(x-2)=0



We calculate terms in parentheses: -((-10x^2+25x+4x-10)), so:

(-10x^2+25x+4x-10)

We get rid of parentheses

-10x^2+25x+4x-10

We add all the numbers together, and all the variables

-10x^2+29x-10

Back to the equation:

-(-10x^2+29x-10)



We get rid of parentheses

10x^2-29x+(x-3)^3-(x^2-3x+5)(x-2)+10=0

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

10x^2-29x+(x-3)^3-(x^2-3x+5)(x-2)=-10

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