4(z-5)+25=29

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Solution for 4(z-5)+25=29 equation: 



4(z-5)+25=29

We move all terms to the left:

4(z-5)+25-(29)=0

We add all the numbers together, and all the variables

4(z-5)-4=0

We multiply parentheses

4z-20-4=0

We add all the numbers together, and all the variables

4z-24=0

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

4z=24

z=24/4

z=6

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