20+4b=1/3b+5.33

Solution for 20+4b=1/3b+5.33 equation:

20+4b=1/3b+5.33

We move all terms to the left:

20+4b-(1/3b+5.33)=0


Domain of the equation: 3b+5.33)!=0

b∈R


We get rid of parentheses

4b-1/3b-5.33+20=0

We multiply all the terms by the denominator


4b*3b-(5.33)*3b+20*3b-1=0

We multiply parentheses

4b*3b-15.99b+20*3b-1=0

Wy multiply elements

12b^2-15.99b+60b-1=0

We add all the numbers together, and all the variables

12b^2+44.01b-1=0

a = 12; b = 44.01; c = -1;
Δ = b2-4ac
Δ = 44.012-4·12·(-1)
Δ = 1984.8801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44.01)-\sqrt{1984.8801}}{2*12}=\frac{-44.01-\sqrt{1984.8801}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44.01)+\sqrt{1984.8801}}{2*12}=\frac{-44.01+\sqrt{1984.8801}}{24}
$