3(y+3)y+1+2=3(y+1)y+1

Solution for 3(y+3)y+1+2=3(y+1)y+1 equation:

3(y+3)y+1+2=3(y+1)y+1

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3y^2+9y-(3(y+1)y+1)+3=0


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

3(y+1)y+1

We multiply parentheses

3y^2+3y+1

Back to the equation:

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


We get rid of parentheses

3y^2-3y^2+9y-3y-1+3=0

We add all the numbers together, and all the variables

6y+2=0

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

6y=-2

y=-2/6

y=-1/3