3(n+4)=1/25n+4

Solution for 3(n+4)=1/25n+4 equation:

3(n+4)=1/25n+4

We move all terms to the left:

3(n+4)-(1/25n+4)=0


Domain of the equation: 25n+4)!=0

n∈R


We multiply parentheses

3n-(1/25n+4)+12=0

We get rid of parentheses

3n-1/25n-4+12=0

We multiply all the terms by the denominator


3n*25n-4*25n+12*25n-1=0

Wy multiply elements

75n^2-100n+300n-1=0

We add all the numbers together, and all the variables

75n^2+200n-1=0

a = 75; b = 200; c = -1;
Δ = b2-4ac
Δ = 2002-4·75·(-1)
Δ = 40300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40300}=\sqrt{100*403}=\sqrt{100}*\sqrt{403}=10\sqrt{403}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-10\sqrt{403}}{2*75}=\frac{-200-10\sqrt{403}}{150}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+10\sqrt{403}}{2*75}=\frac{-200+10\sqrt{403}}{150}
$