(2x-4)(3x+1)=(6x-5)(x+2)

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Solution for (2x-4)(3x+1)=(6x-5)(x+2) equation: 



(2x-4)(3x+1)=(6x-5)(x+2)

We move all terms to the left:

(2x-4)(3x+1)-((6x-5)(x+2))=0

We multiply parentheses ..

(+6x^2+2x-12x-4)-((6x-5)(x+2))=0



We calculate terms in parentheses: -((6x-5)(x+2)), so:

(6x-5)(x+2)

We multiply parentheses ..

(+6x^2+12x-5x-10)

We get rid of parentheses

6x^2+12x-5x-10

We add all the numbers together, and all the variables

6x^2+7x-10

Back to the equation:

-(6x^2+7x-10)



We get rid of parentheses

6x^2-6x^2+2x-12x-7x-4+10=0

We add all the numbers together, and all the variables

-17x+6=0

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

-17x=-6

x=-6/-17

x=6/17

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