2/3b+b=20-b

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

2/3b+b=20-b

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

6b^2-60b+2=0

a = 6; b = -60; c = +2;
Δ = b2-4ac
Δ = -602-4·6·2
Δ = 3552
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:

$\sqrt{\Delta}=\sqrt{3552}=\sqrt{16*222}=\sqrt{16}*\sqrt{222}=4\sqrt{222}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{222}}{2*6}=\frac{60-4\sqrt{222}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{222}}{2*6}=\frac{60+4\sqrt{222}}{12}
$