3/3b+4=2/b-2

Solution for 3/3b+4=2/b-2 equation:

3/3b+4=2/b-2

We move all terms to the left:

3/3b+4-(2/b-2)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: b-2)!=0

b∈R


We get rid of parentheses

3/3b-2/b+2+4=0

We calculate fractions


3b/3b^2+(-6b)/3b^2+2+4=0

We add all the numbers together, and all the variables

3b/3b^2+(-6b)/3b^2+6=0

We multiply all the terms by the denominator


3b+(-6b)+6*3b^2=0

Wy multiply elements

18b^2+3b+(-6b)=0

We get rid of parentheses

18b^2+3b-6b=0

We add all the numbers together, and all the variables

18b^2-3b=0

a = 18; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·18·0
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*18}=\frac{0}{36}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*18}=\frac{6}{36}
=1/6
$