-12k2-6k+11=0

Solution for -12k2-6k+11=0 equation:

-12k^2-6k+11=0

a = -12; b = -6; c = +11;
Δ = b2-4ac
Δ = -62-4·(-12)·11
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{-(-6)-2\sqrt{141}}{2*-12}=\frac{6-2\sqrt{141}}{-24}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{141}}{2*-12}=\frac{6+2\sqrt{141}}{-24}
$