10(z+4)-4(z-2)=-2-(6z-2)

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Solution for 10(z+4)-4(z-2)=-2-(6z-2) equation: 



10(z+4)-4(z-2)=-2-(6z-2)

We move all terms to the left:

10(z+4)-4(z-2)-(-2-(6z-2))=0

We multiply parentheses

10z-4z-(-2-(6z-2))+40+8=0



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

-2-(6z-2)

determiningTheFunctionDomain
-(6z-2)-2

We get rid of parentheses

-6z+2-2

We add all the numbers together, and all the variables

-6z

Back to the equation:

-(-6z)



We add all the numbers together, and all the variables

6z-(-6z)+48=0

We get rid of parentheses

6z+6z+48=0

We add all the numbers together, and all the variables

12z+48=0

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

12z=-48

z=-48/12

z=-4

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