r^2-24r+72=0

Solution for r^2-24r+72=0 equation:

r^2-24r+72=0

a = 1; b = -24; c = +72;
Δ = b2-4ac
Δ = -242-4·1·72
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{2}}{2*1}=\frac{24-12\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{2}}{2*1}=\frac{24+12\sqrt{2}}{2}
$