12/4n-18=n-4

Solution for 12/4n-18=n-4 equation:

12/4n-18=n-4

We move all terms to the left:

12/4n-18-(n-4)=0


Domain of the equation: 4n!=0

n!=0/4

n!=0

n∈R


We get rid of parentheses

12/4n-n+4-18=0

We multiply all the terms by the denominator


-n*4n+4*4n-18*4n+12=0

Wy multiply elements

-4n^2+16n-72n+12=0

We add all the numbers together, and all the variables

-4n^2-56n+12=0

a = -4; b = -56; c = +12;
Δ = b2-4ac
Δ = -562-4·(-4)·12
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-16\sqrt{13}}{2*-4}=\frac{56-16\sqrt{13}}{-8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+16\sqrt{13}}{2*-4}=\frac{56+16\sqrt{13}}{-8}
$