20+4b=1/3b+16

Solution for 20+4b=1/3b+16 equation:

20+4b=1/3b+16

We move all terms to the left:

20+4b-(1/3b+16)=0


Domain of the equation: 3b+16)!=0

b∈R


We get rid of parentheses

4b-1/3b-16+20=0

We multiply all the terms by the denominator


4b*3b-16*3b+20*3b-1=0

Wy multiply elements

12b^2-48b+60b-1=0

We add all the numbers together, and all the variables

12b^2+12b-1=0

a = 12; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·12·(-1)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*12}=\frac{-12-8\sqrt{3}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*12}=\frac{-12+8\sqrt{3}}{24}
$