16=1/2(4(3+b2)

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Solution for 16=1/2(4(3+b2) equation: 



16=1/2(4(3+b2)

We move all terms to the left:

16-(1/2(4(3+b2))=0



Domain of the equation: 2(4(3+b2))+16!=0

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

2(4(3+b2))!=-16

b∈R



We add all the numbers together, and all the variables

-(1/2(4(+b^2+3))+16=0

We multiply all the terms by the denominator


-(1=0

We add all the numbers together, and all the variables

=0

b=0/1

b=0

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