24k^2+108k+108=0

Solution for 24k^2+108k+108=0 equation:

24k^2+108k+108=0

a = 24; b = 108; c = +108;
Δ = b2-4ac
Δ = 1082-4·24·108
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(108)-36}{2*24}=\frac{-144}{48}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(108)+36}{2*24}=\frac{-72}{48}
=-1+1/2
$