1/3b+4b=5

Solution for 1/3b+4b=5 equation:

1/3b+4b=5

We move all terms to the left:

1/3b+4b-(5)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

4b+1/3b-5=0

We multiply all the terms by the denominator


4b*3b-5*3b+1=0

Wy multiply elements

12b^2-15b+1=0

a = 12; b = -15; c = +1;
Δ = b2-4ac
Δ = -152-4·12·1
Δ = 177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{177}}{2*12}=\frac{15-\sqrt{177}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{177}}{2*12}=\frac{15+\sqrt{177}}{24}
$