r^2+24r-576=0

Solution for r^2+24r-576=0 equation:

r^2+24r-576=0

a = 1; b = 24; c = -576;
Δ = b2-4ac
Δ = 242-4·1·(-576)
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{5}}{2*1}=\frac{-24-24\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{5}}{2*1}=\frac{-24+24\sqrt{5}}{2}
$