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

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



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

We move all terms to the left:

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

We multiply parentheses

12z-4z-(3(z-2)+4(z-2))+24+16=0



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

3(z-2)+4(z-2)

We multiply parentheses

3z+4z-6-8

We add all the numbers together, and all the variables

7z-14

Back to the equation:

-(7z-14)



We add all the numbers together, and all the variables

8z-(7z-14)+40=0

We get rid of parentheses

8z-7z+14+40=0

We add all the numbers together, and all the variables

z+54=0

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

z=-54

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