(4/n-6)+(4n/8n-48)=0

Solution for (4/n-6)+(4n/8n-48)=0 equation:

(4/n-6)+(4n/8n-48)=0


Domain of the equation: n-6)!=0

n∈R



Domain of the equation: 8n-48)!=0

n∈R


We get rid of parentheses

4/n+4n/8n-6-48=0

We calculate fractions


4n^2/8n^2+32n/8n^2-6-48=0

We add all the numbers together, and all the variables

4n^2/8n^2+32n/8n^2-54=0

We multiply all the terms by the denominator


4n^2+32n-54*8n^2=0

Wy multiply elements

4n^2-432n^2+32n=0

We add all the numbers together, and all the variables

-428n^2+32n=0

a = -428; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-428)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-428}=\frac{-64}{-856}
=8/107
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-428}=\frac{0}{-856}
=0
$