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 = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{6}}{2*1}=\frac{-24-12\sqrt{6}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{6}}{2*1}=\frac{-24+12\sqrt{6}}{2}
$