3-(40/3b)=3b/b+4

Solution for 3-(40/3b)=3b/b+4 equation:

3-(40/3b)=3b/b+4

We move all terms to the left:

3-(40/3b)-(3b/b+4)=0


Domain of the equation: 3b)!=0

b!=0/1

b!=0

b∈R



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

b∈R


We add all the numbers together, and all the variables

-(+40/3b)-(3b/b+4)+3=0

We get rid of parentheses

-40/3b-3b/b-4+3=0

We calculate fractions


(-9b^2)/3b^2+(-40b)/3b^2-4+3=0

We add all the numbers together, and all the variables

(-9b^2)/3b^2+(-40b)/3b^2-1=0

We multiply all the terms by the denominator


(-9b^2)+(-40b)-1*3b^2=0

Wy multiply elements

(-9b^2)-3b^2+(-40b)=0

We get rid of parentheses

-9b^2-3b^2-40b=0

We add all the numbers together, and all the variables

-12b^2-40b=0

a = -12; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-12)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-12}=\frac{0}{-24}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-12}=\frac{80}{-24}
=-3+1/3
$