n^2+18n=72

Solution for n^2+18n=72 equation:

n^2+18n=72

We move all terms to the left:

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}
$