2/3b+b=20

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

2/3b+b=20

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

b+2/3b-20=0

We multiply all the terms by the denominator


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

Wy multiply elements

3b^2-60b+2=0

a = 3; b = -60; c = +2;
Δ = b2-4ac
Δ = -602-4·3·2
Δ = 3576
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:

$\sqrt{\Delta}=\sqrt{3576}=\sqrt{4*894}=\sqrt{4}*\sqrt{894}=2\sqrt{894}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-2\sqrt{894}}{2*3}=\frac{60-2\sqrt{894}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+2\sqrt{894}}{2*3}=\frac{60+2\sqrt{894}}{6}
$