6(3z+2)+1=23(z-4)

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Solution for 6(3z+2)+1=23(z-4) equation: 



6(3z+2)+1=23(z-4)

We move all terms to the left:

6(3z+2)+1-(23(z-4))=0

We multiply parentheses

18z-(23(z-4))+12+1=0



We calculate terms in parentheses: -(23(z-4)), so:

23(z-4)

We multiply parentheses

23z-92

Back to the equation:

-(23z-92)



We add all the numbers together, and all the variables

18z-(23z-92)+13=0

We get rid of parentheses

18z-23z+92+13=0

We add all the numbers together, and all the variables

-5z+105=0

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

-5z=-105

z=-105/-5

z=+21

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