p^2-18p=120

Solution for p^2-18p=120 equation:

p^2-18p=120

We move all terms to the left:

p^2-18p-(120)=0

a = 1; b = -18; c = -120;
Δ = b2-4ac
Δ = -182-4·1·(-120)
Δ = 804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{201}}{2*1}=\frac{18-2\sqrt{201}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{201}}{2*1}=\frac{18+2\sqrt{201}}{2}
$