3/2b-13/(2(b+2))=6

Solution for 3/2b-13/(2(b+2))=6 equation:

3/2b-13/(2(b+2))=6

We move all terms to the left:

3/2b-13/(2(b+2))-(6)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R



Domain of the equation: (2(b+2))!=0

b∈R


We calculate fractions


(3*(2(b+2)))/(2b*(2(b+2)))+(-26b)/(2b*(2(b+2)))-6=0


We calculate terms in parentheses: +(3*(2(b+2)))/(2b*(2(b+2))), so:

3*(2(b+2)))/(2b*(2(b+2))

We multiply all the terms by the denominator


3*(2(b+2)))

Back to the equation:

+(3*(2(b+2))))



We calculate terms in parentheses: +(-26b)/(2b*(2(b+2))), so:

-26b)/(2b*(2(b+2))

We multiply all the terms by the denominator


-26b)

Back to the equation:

+(-26b))