4y(y+3)-4=4(y+1)(y+3)

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

4y(y+3)-4=4(y+1)(y+3)

We move all terms to the left:

4y(y+3)-4-(4(y+1)(y+3))=0

We multiply parentheses

4y^2+12y-(4(y+1)(y+3))-4=0

We multiply parentheses ..

4y^2-(4(+y^2+3y+y+3))+12y-4=0


We calculate terms in parentheses: -(4(+y^2+3y+y+3)), so:

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

We multiply parentheses

4y^2+12y+4y+12

We add all the numbers together, and all the variables

4y^2+16y+12

Back to the equation:

-(4y^2+16y+12)


We add all the numbers together, and all the variables

4y^2+12y-(4y^2+16y+12)-4=0

We get rid of parentheses

4y^2-4y^2+12y-16y-12-4=0

We add all the numbers together, and all the variables

-4y-16=0

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

-4y=16

y=16/-4

y=-4