36k^2+6k-240=0

Solution for 36k^2+6k-240=0 equation:

36k^2+6k-240=0

a = 36; b = 6; c = -240;
Δ = b2-4ac
Δ = 62-4·36·(-240)
Δ = 34596
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{34596}=186$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-186}{2*36}=\frac{-192}{72}
=-2+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+186}{2*36}=\frac{180}{72}
=2+1/2
$