2/3b+5=20-b*2

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

2/3b+5=20-b*2

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

We multiply parentheses

6b^2-20*3b+5*3b+2=0

Wy multiply elements

6b^2-60b+15b+2=0

We add all the numbers together, and all the variables

6b^2-45b+2=0

a = 6; b = -45; c = +2;
Δ = b2-4ac
Δ = -452-4·6·2
Δ = 1977
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{1977}}{2*6}=\frac{45-\sqrt{1977}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1977}}{2*6}=\frac{45+\sqrt{1977}}{12}
$