3k^2+75k+24=0

Solution for 3k^2+75k+24=0 equation:

3k^2+75k+24=0

a = 3; b = 75; c = +24;
Δ = b2-4ac
Δ = 752-4·3·24
Δ = 5337
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:

$\sqrt{\Delta}=\sqrt{5337}=\sqrt{9*593}=\sqrt{9}*\sqrt{593}=3\sqrt{593}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(75)-3\sqrt{593}}{2*3}=\frac{-75-3\sqrt{593}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+3\sqrt{593}}{2*3}=\frac{-75+3\sqrt{593}}{6}
$