16=1/2.4(3+b2)

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))!=0

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+16*2.4(+b^2+3))=0


We calculate terms in parentheses: -(1+16*2.4(+b^2+3)), so:

1+16*2.4(+b^2+3)

determiningTheFunctionDomain
16*2.4(+b^2+3)+1

Wy multiply elements

32b+1

Back to the equation:

-(32b+1)


We get rid of parentheses

-32b-1=0

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

-32b=1

b=1/-32

b=-1/32