6k^2+69k+192=0

Solution for 6k^2+69k+192=0 equation:

6k^2+69k+192=0

a = 6; b = 69; c = +192;
Δ = b2-4ac
Δ = 692-4·6·192
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-3\sqrt{17}}{2*6}=\frac{-69-3\sqrt{17}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+3\sqrt{17}}{2*6}=\frac{-69+3\sqrt{17}}{12}
$