4(3-z)-22=16z-5(2+4z)

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Solution for 4(3-z)-22=16z-5(2+4z) equation: 



4(3-z)-22=16z-5(2+4z)

We move all terms to the left:

4(3-z)-22-(16z-5(2+4z))=0

We add all the numbers together, and all the variables

4(-1z+3)-(16z-5(4z+2))-22=0

We multiply parentheses

-4z-(16z-5(4z+2))+12-22=0



We calculate terms in parentheses: -(16z-5(4z+2)), so:

16z-5(4z+2)

We multiply parentheses

16z-20z-10

We add all the numbers together, and all the variables

-4z-10

Back to the equation:

-(-4z-10)



We add all the numbers together, and all the variables

-4z-(-4z-10)-10=0

We get rid of parentheses

-4z+4z+10-10=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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