-3(2+6z)=4(z+1)+12

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Solution for -3(2+6z)=4(z+1)+12 equation: 



-3(2+6z)=4(z+1)+12

We move all terms to the left:

-3(2+6z)-(4(z+1)+12)=0

We add all the numbers together, and all the variables

-3(6z+2)-(4(z+1)+12)=0

We multiply parentheses

-18z-(4(z+1)+12)-6=0



We calculate terms in parentheses: -(4(z+1)+12), so:

4(z+1)+12

We multiply parentheses

4z+4+12

We add all the numbers together, and all the variables

4z+16

Back to the equation:

-(4z+16)



We get rid of parentheses

-18z-4z-16-6=0

We add all the numbers together, and all the variables

-22z-22=0

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

-22z=22

z=22/-22

z=-1

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