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

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

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

We move all terms to the left:

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


Domain of the equation: 2)4(3+b^2))!=0

b∈R


We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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


We calculate terms in parentheses: -((+1+16*2)4(3+b^2)), so:

(+1+16*2)4(3+b^2)

We add all the numbers together, and all the variables

334(3+b^2)

We multiply parentheses

334b^2+1002

Back to the equation:

-(334b^2+1002)


We get rid of parentheses

-334b^2-1002=0

a = -334; b = 0; c = -1002;
Δ = b2-4ac
Δ = 02-4·(-334)·(-1002)
Δ = -1338672
Delta is less than zero, so there is no solution for the equation