3k^2-24k-48=0

Solution for 3k^2-24k-48=0 equation:

3k^2-24k-48=0

a = 3; b = -24; c = -48;
Δ = b2-4ac
Δ = -242-4·3·(-48)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24\sqrt{2}}{2*3}=\frac{24-24\sqrt{2}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24\sqrt{2}}{2*3}=\frac{24+24\sqrt{2}}{6}
$