b/3b+4=2b+6b-16

Solution for b/3b+4=2b+6b-16 equation:

b/3b+4=2b+6b-16

We move all terms to the left:

b/3b+4-(2b+6b-16)=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/3b-(8b-16)+4=0

We get rid of parentheses

b/3b-8b+16+4=0

We multiply all the terms by the denominator


b-8b*3b+16*3b+4*3b=0

Wy multiply elements

-24b^2+b+48b+12b=0

We add all the numbers together, and all the variables

-24b^2+61b=0

a = -24; b = 61; c = 0;
Δ = b2-4ac
Δ = 612-4·(-24)·0
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(61)-61}{2*-24}=\frac{-122}{-48}
=2+13/24
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(61)+61}{2*-24}=\frac{0}{-48}
=0
$