n(n-18)=72

Solution for n(n-18)=72 equation:

n(n-18)=72

We move all terms to the left:

n(n-18)-(72)=0

We multiply parentheses

n^2-18n-72=0

a = 1; b = -18; c = -72;
Δ = b2-4ac
Δ = -182-4·1·(-72)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{17}}{2*1}=\frac{18-6\sqrt{17}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{17}}{2*1}=\frac{18+6\sqrt{17}}{2}
$