n(n+1)/(n-3)=72/5

Solution for n(n+1)/(n-3)=72/5 equation:

n(n+1)/(n-3)=72/5

We move all terms to the left:

n(n+1)/(n-3)-(72/5)=0


Domain of the equation: (n-3)!=0

We move all terms containing n to the left, all other terms to the right

n!=3

n∈R


We add all the numbers together, and all the variables

n(n+1)/(n-3)-(+72/5)=0

We get rid of parentheses

n(n+1)/(n-3)-72/5=0

We calculate fractions


(5n^2+5n)/(5n-15)+(-72n+216)/(5n-15)=0

We multiply all the terms by the denominator


(5n^2+5n)+(-72n+216)=0

We get rid of parentheses

5n^2+5n-72n+216=0

We add all the numbers together, and all the variables

5n^2-67n+216=0

a = 5; b = -67; c = +216;
Δ = b2-4ac
Δ = -672-4·5·216
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-67)-13}{2*5}=\frac{54}{10}
=5+2/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-67)+13}{2*5}=\frac{80}{10}
=8
$