5(3b-4)+6(2b+3)+12=64

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Solution for 5(3b-4)+6(2b+3)+12=64 equation: 



5(3b-4)+6(2b+3)+12=64

We move all terms to the left:

5(3b-4)+6(2b+3)+12-(64)=0

We add all the numbers together, and all the variables

5(3b-4)+6(2b+3)-52=0

We multiply parentheses

15b+12b-20+18-52=0

We add all the numbers together, and all the variables

27b-54=0

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

27b=54

b=54/27

b=2

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