48-5(b-4)=2(6b+2)+20

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Solution for 48-5(b-4)=2(6b+2)+20 equation: 



48-5(b-4)=2(6b+2)+20

We move all terms to the left:

48-5(b-4)-(2(6b+2)+20)=0

We multiply parentheses

-5b-(2(6b+2)+20)+20+48=0



We calculate terms in parentheses: -(2(6b+2)+20), so:

2(6b+2)+20

We multiply parentheses

12b+4+20

We add all the numbers together, and all the variables

12b+24

Back to the equation:

-(12b+24)



We add all the numbers together, and all the variables

-5b-(12b+24)+68=0

We get rid of parentheses

-5b-12b-24+68=0

We add all the numbers together, and all the variables

-17b+44=0

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

-17b=-44

b=-44/-17

b=2+10/17

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