2/9b=1/3b-10

Solution for 2/9b=1/3b-10 equation:

2/9b=1/3b-10

We move all terms to the left:

2/9b-(1/3b-10)=0


Domain of the equation: 9b!=0

b!=0/9

b!=0

b∈R



Domain of the equation: 3b-10)!=0

b∈R


We get rid of parentheses

2/9b-1/3b+10=0

We calculate fractions


6b/27b^2+(-9b)/27b^2+10=0

We multiply all the terms by the denominator


6b+(-9b)+10*27b^2=0

Wy multiply elements

270b^2+6b+(-9b)=0

We get rid of parentheses

270b^2+6b-9b=0

We add all the numbers together, and all the variables

270b^2-3b=0

a = 270; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·270·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*270}=\frac{0}{540}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*270}=\frac{6}{540}
=1/90
$