12(z+4)-4(z-2)=5(z-2)+2(z-2)

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Solution for 12(z+4)-4(z-2)=5(z-2)+2(z-2) equation: 



12(z+4)-4(z-2)=5(z-2)+2(z-2)

We move all terms to the left:

12(z+4)-4(z-2)-(5(z-2)+2(z-2))=0

We multiply parentheses

12z-4z-(5(z-2)+2(z-2))+48+8=0



We calculate terms in parentheses: -(5(z-2)+2(z-2)), so:

5(z-2)+2(z-2)

We multiply parentheses

5z+2z-10-4

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)+56=0

We get rid of parentheses

8z-7z+14+56=0

We add all the numbers together, and all the variables

z+70=0

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

z=-70

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