(3x+4)(3x+2)(x-1/3)(x-1)=36

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Solution for (3x+4)(3x+2)(x-1/3)(x-1)=36 equation: 



(3x+4)(3x+2)(x-1/3)(x-1)=36

We move all terms to the left:

(3x+4)(3x+2)(x-1/3)(x-1)-(36)=0



Domain of the equation: 3)(x-1)!=0

x∈R



We add all the numbers together, and all the variables

(3x+4)(3x+2)(+x-1/3)(x-1)-36=0

We multiply parentheses ..

(+9x^2+6x+12x+8)(+x-1/3)(x-1)-36=0

We multiply all the terms by the denominator


(+9x^2+6x+12x+8)(+x-1-36*3)(x-1)=0

We add all the numbers together, and all the variables

(+9x^2+6x+12x+8)(x-109)(x-1)=0

We multiply parentheses ..

(+9x^2+6x+12x+8)(+x^2-1x-109x+109)=0

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