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

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



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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2-3x+2x-1)+(2x+1)(3x-2)-(12(x-3)(x+1))=0



We calculate terms in parentheses: -(12(x-3)(x+1)), so:

12(x-3)(x+1)

We multiply parentheses ..

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

We multiply parentheses

12x^2+12x-36x-36

We add all the numbers together, and all the variables

12x^2-24x-36

Back to the equation:

-(12x^2-24x-36)



We get rid of parentheses

6x^2-12x^2-3x+2x+(2x+1)(3x-2)+24x-1+36=0

We multiply parentheses ..

6x^2-12x^2+(+6x^2-4x+3x-2)-3x+2x+24x-1+36=0

We add all the numbers together, and all the variables

-6x^2+(+6x^2-4x+3x-2)+23x+35=0

We get rid of parentheses

-6x^2+6x^2-4x+3x+23x-2+35=0

We add all the numbers together, and all the variables

22x+33=0

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

22x=-33

x=-33/22

x=-1+1/2

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