c^2+24c+36=0

Solution for c^2+24c+36=0 equation:

c^2+24c+36=0

a = 1; b = 24; c = +36;
Δ = b2-4ac
Δ = 242-4·1·36
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{3}}{2*1}=\frac{-24-12\sqrt{3}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{3}}{2*1}=\frac{-24+12\sqrt{3}}{2}
$