19b=(1/2b)(b-3)

Solution for 19b=(1/2b)(b-3) equation:

19b=(1/2b)(b-3)

We move all terms to the left:

19b-((1/2b)(b-3))=0


Domain of the equation: 2b)(b-3))!=0

b∈R


We add all the numbers together, and all the variables

19b-((+1/2b)(b-3))=0

We multiply parentheses ..

-((+b^2-3b))+19b=0


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

(+b^2-3b)

We get rid of parentheses

b^2-3b

Back to the equation:

-(b^2-3b)


We add all the numbers together, and all the variables

19b-(b^2-3b)=0

We get rid of parentheses

-b^2+19b+3b=0

We add all the numbers together, and all the variables

-1b^2+22b=0

a = -1; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-1)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-1}=\frac{-44}{-2}
=+22
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-1}=\frac{0}{-2}
=0
$