20-b=2/3b+5

Solution for 20-b=2/3b+5 equation:

20-b=2/3b+5

We move all terms to the left:

20-b-(2/3b+5)=0


Domain of the equation: 3b+5)!=0

b∈R


We add all the numbers together, and all the variables

-1b-(2/3b+5)+20=0

We get rid of parentheses

-1b-2/3b-5+20=0

We multiply all the terms by the denominator


-1b*3b-5*3b+20*3b-2=0

Wy multiply elements

-3b^2-15b+60b-2=0

We add all the numbers together, and all the variables

-3b^2+45b-2=0

a = -3; b = 45; c = -2;
Δ = b2-4ac
Δ = 452-4·(-3)·(-2)
Δ = 2001
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{-(45)-\sqrt{2001}}{2*-3}=\frac{-45-\sqrt{2001}}{-6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2001}}{2*-3}=\frac{-45+\sqrt{2001}}{-6}
$