r^2+12r-264=0

Solution for r^2+12r-264=0 equation:

r^2+12r-264=0

a = 1; b = 12; c = -264;
Δ = b2-4ac
Δ = 122-4·1·(-264)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20\sqrt{3}}{2*1}=\frac{-12-20\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20\sqrt{3}}{2*1}=\frac{-12+20\sqrt{3}}{2}
$