12+3z+4z=-4(3z-6)+26

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Solution for 12+3z+4z=-4(3z-6)+26 equation: 



12+3z+4z=-4(3z-6)+26

We move all terms to the left:

12+3z+4z-(-4(3z-6)+26)=0

We add all the numbers together, and all the variables

7z-(-4(3z-6)+26)+12=0



We calculate terms in parentheses: -(-4(3z-6)+26), so:

-4(3z-6)+26

We multiply parentheses

-12z+24+26

We add all the numbers together, and all the variables

-12z+50

Back to the equation:

-(-12z+50)



We get rid of parentheses

7z+12z-50+12=0

We add all the numbers together, and all the variables

19z-38=0

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

19z=38

z=38/19

z=2

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