6+2/3b=2/4b-4

Solution for 6+2/3b=2/4b-4 equation:

6+2/3b=2/4b-4

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 4b-4)!=0

b∈R


We get rid of parentheses

2/3b-2/4b+4+6=0

We calculate fractions


8b/12b^2+(-6b)/12b^2+4+6=0

We add all the numbers together, and all the variables

8b/12b^2+(-6b)/12b^2+10=0

We multiply all the terms by the denominator


8b+(-6b)+10*12b^2=0

Wy multiply elements

120b^2+8b+(-6b)=0

We get rid of parentheses

120b^2+8b-6b=0

We add all the numbers together, and all the variables

120b^2+2b=0

a = 120; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·120·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*120}=\frac{-4}{240}
=-1/60
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*120}=\frac{0}{240}
=0
$