8(4b+3)+8=128

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Solution for 8(4b+3)+8=128 equation: 



8(4b+3)+8=128

We move all terms to the left:

8(4b+3)+8-(128)=0

We add all the numbers together, and all the variables

8(4b+3)-120=0

We multiply parentheses

32b+24-120=0

We add all the numbers together, and all the variables

32b-96=0

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

32b=96

b=96/32

b=3

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