B=(8/3)(j-3-)

Solution for B=(8/3)(j-3-) equation:

=(8/3)(B-3-)

We move all terms to the left:

-((8/3)(B-3-))=0


Domain of the equation: 3)(B-3-))!=0

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

3)(B-))!=3

B∈R


We add all the numbers together, and all the variables

-((+8/3)(+B))=0

We multiply parentheses ..

-((+8B^2))=0


We calculate terms in parentheses: -((+8B^2)), so:

(+8B^2)

We get rid of parentheses

8B^2

Back to the equation:

-(8B^2)


a = -8; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-8)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$B=\frac{-b}{2a}=\frac{0}{-16}=0$