(2y-9)(3y+1)=(6y+4)(y-2)

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Solution for (2y-9)(3y+1)=(6y+4)(y-2) equation: 



(2y-9)(3y+1)=(6y+4)(y-2)

We move all terms to the left:

(2y-9)(3y+1)-((6y+4)(y-2))=0

We multiply parentheses ..

(+6y^2+2y-27y-9)-((6y+4)(y-2))=0



We calculate terms in parentheses: -((6y+4)(y-2)), so:

(6y+4)(y-2)

We multiply parentheses ..

(+6y^2-12y+4y-8)

We get rid of parentheses

6y^2-12y+4y-8

We add all the numbers together, and all the variables

6y^2-8y-8

Back to the equation:

-(6y^2-8y-8)



We get rid of parentheses

6y^2-6y^2+2y-27y+8y-9+8=0

We add all the numbers together, and all the variables

-17y-1=0

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

-17y=1

y=1/-17

y=-1/17

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