3k^2+2k-36=0

Solution for 3k^2+2k-36=0 equation:

3k^2+2k-36=0

a = 3; b = 2; c = -36;
Δ = b2-4ac
Δ = 22-4·3·(-36)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{109}}{2*3}=\frac{-2-2\sqrt{109}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{109}}{2*3}=\frac{-2+2\sqrt{109}}{6}
$