B+2(b+3)+4b-5=148

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Solution for B+2(b+3)+4b-5=148 equation: 



+2(B+3)+4B-5=148

We move all terms to the left:

+2(B+3)+4B-5-(148)=0

We add all the numbers together, and all the variables

4B+2(B+3)-153=0

We multiply parentheses

4B+2B+6-153=0

We add all the numbers together, and all the variables

6B-147=0

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

6B=147

B=147/6

B=24+1/2

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