(x^2-8x-2)/3=(x^2-3x+2)/2

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Solution for (x^2-8x-2)/3=(x^2-3x+2)/2 equation: 



(x^2-8x-2)/3=(x^2-3x+2)/2

We move all terms to the left:

(x^2-8x-2)/3-((x^2-3x+2)/2)=0

We calculate fractions


(x^2-8x)/()+(-((x^2-3x+2)*3)/()=0



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

-((x^2-3x+2)*3)/(

We multiply all the terms by the denominator


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



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

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

We multiply parentheses

3x^2-9x+6

Back to the equation:

-(3x^2-9x+6)



We get rid of parentheses

-3x^2+9x-6

Back to the equation:

+(-3x^2+9x-6)



We get rid of parentheses

-3x^2+9x+(x^2-8x)/()-6=0

We multiply all the terms by the denominator


-3x^2*()+9x*()+(x^2-8x)-6*()=0

We add all the numbers together, and all the variables

-3x^2*()+9x*()+(x^2-8x)=0

We get rid of parentheses

-3x^2*()+x^2+9x*()-8x=0

We add all the numbers together, and all the variables

x^2-3x^2*()-8x+9x*()=0

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