20=(z+2)+(2z-6)

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Solution for 20=(z+2)+(2z-6) equation: 



20=(z+2)+(2z-6)

We move all terms to the left:

20-((z+2)+(2z-6))=0



We calculate terms in parentheses: -((z+2)+(2z-6)), so:

(z+2)+(2z-6)

We get rid of parentheses

z+2z+2-6

We add all the numbers together, and all the variables

3z-4

Back to the equation:

-(3z-4)



We get rid of parentheses

-3z+4+20=0

We add all the numbers together, and all the variables

-3z+24=0

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

-3z=-24

z=-24/-3

z=+8

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