3/3b+4=2b-4

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

3/3b+4=2b-4

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-6b^2+12b+12b+3=0

We add all the numbers together, and all the variables

-6b^2+24b+3=0

a = -6; b = 24; c = +3;
Δ = b2-4ac
Δ = 242-4·(-6)·3
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-18\sqrt{2}}{2*-6}=\frac{-24-18\sqrt{2}}{-12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+18\sqrt{2}}{2*-6}=\frac{-24+18\sqrt{2}}{-12}
$