3(2z+12)=4(z-1)+104

Solution for 3(2z+12)=4(z-1)+104 equation:

3(2z+12)=4(z-1)+104

We move all terms to the left:

3(2z+12)-(4(z-1)+104)=0

We multiply parentheses

6z-(4(z-1)+104)+36=0


We calculate terms in parentheses: -(4(z-1)+104), so:

4(z-1)+104

We multiply parentheses

4z-4+104

We add all the numbers together, and all the variables

4z+100

Back to the equation:

-(4z+100)


We get rid of parentheses

6z-4z-100+36=0

We add all the numbers together, and all the variables

2z-64=0

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

2z=64

z=64/2

z=32