n(n-3)/2=4.n

Solution for n(n-3)/2=4.n equation:

n(n-3)/2=4.n

We move all terms to the left:

n(n-3)/2-(4.n)=0

We add all the numbers together, and all the variables

n(n-3)/2-(+4.n)=0

We get rid of parentheses

n(n-3)/2-4.n=0

We multiply all the terms by the denominator


n(n-3)-(4.n)*2=0

We add all the numbers together, and all the variables

n(n-3)-(+4.n)*2=0

We multiply parentheses

n^2-3n-8n=0

We add all the numbers together, and all the variables

n^2-11n=0

a = 1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·1·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*1}=\frac{0}{2}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*1}=\frac{22}{2}
=11
$