r^2-36r+288=0

Solution for r^2-36r+288=0 equation:

r^2-36r+288=0

a = 1; b = -36; c = +288;
Δ = b2-4ac
Δ = -362-4·1·288
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*1}=\frac{24}{2}
=12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*1}=\frac{48}{2}
=24
$