5k^2+24k-18=0

Solution for 5k^2+24k-18=0 equation:

5k^2+24k-18=0

a = 5; b = 24; c = -18;
Δ = b2-4ac
Δ = 242-4·5·(-18)
Δ = 936
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:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{26}}{2*5}=\frac{-24-6\sqrt{26}}{10}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{26}}{2*5}=\frac{-24+6\sqrt{26}}{10}
$