2/3b+4/3b+b=15

Solution for 2/3b+4/3b+b=15 equation:

2/3b+4/3b+b=15

We move all terms to the left:

2/3b+4/3b+b-(15)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

b+2/3b+4/3b-15=0

We multiply all the terms by the denominator


b*3b-15*3b+2+4=0

We add all the numbers together, and all the variables

b*3b-15*3b+6=0

Wy multiply elements

3b^2-45b+6=0

a = 3; b = -45; c = +6;
Δ = b2-4ac
Δ = -452-4·3·6
Δ = 1953
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:

$\sqrt{\Delta}=\sqrt{1953}=\sqrt{9*217}=\sqrt{9}*\sqrt{217}=3\sqrt{217}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3\sqrt{217}}{2*3}=\frac{45-3\sqrt{217}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3\sqrt{217}}{2*3}=\frac{45+3\sqrt{217}}{6}
$