((n)(n-1)=72)

Solution for ((n)(n-1)=72) equation:

((n)(n-1)=72)

We move all terms to the left:

((n)(n-1)-(72))=0


We calculate terms in parentheses: +(n(n-1)-72), so:

n(n-1)-72

We multiply parentheses

n^2-1n-72

Back to the equation:

+(n^2-1n-72)


We get rid of parentheses

n^2-1n-72=0

a = 1; b = -1; c = -72;
Δ = b2-4ac
Δ = -12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*1}=\frac{-16}{2}
=-8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*1}=\frac{18}{2}
=9
$