3y(y-1)=2y(y-2)-(3-y2)

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Solution for 3y(y-1)=2y(y-2)-(3-y2) equation: 



3y(y-1)=2y(y-2)-(3-y2)

We move all terms to the left:

3y(y-1)-(2y(y-2)-(3-y2))=0

We add all the numbers together, and all the variables

-(2y(y-2)-(-1y^2+3))+3y(y-1)=0

We multiply parentheses

-(2y(y-2)-(-1y^2+3))+3y^2-3y=0



We calculate terms in parentheses: -(2y(y-2)-(-1y^2+3)), so:

2y(y-2)-(-1y^2+3)

determiningTheFunctionDomain
-(-1y^2+3)+2y(y-2)

We multiply parentheses

-(-1y^2+3)+2y^2-4y

We get rid of parentheses

1y^2+2y^2-4y-3

We add all the numbers together, and all the variables

3y^2-4y-3

Back to the equation:

-(3y^2-4y-3)



We add all the numbers together, and all the variables

3y^2-3y-(3y^2-4y-3)=0

We get rid of parentheses

3y^2-3y^2-3y+4y+3=0

We add all the numbers together, and all the variables

y+3=0

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

y=-3

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