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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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



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

(6x-5)(+x)

We multiply parentheses ..

(+6x^2-5x)

We get rid of parentheses

6x^2-5x

Back to the equation:

-(6x^2-5x)



We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x-4=0

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

-5x=4

x=4/-5

x=-4/5

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