3k2-18k-20=0

Solution for 3k2-18k-20=0 equation:

3k^2-18k-20=0

a = 3; b = -18; c = -20;
Δ = b2-4ac
Δ = -182-4·3·(-20)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{141}}{2*3}=\frac{18-2\sqrt{141}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{141}}{2*3}=\frac{18+2\sqrt{141}}{6}
$