3k^2+22k-1=0

Solution for 3k^2+22k-1=0 equation:

3k^2+22k-1=0

a = 3; b = 22; c = -1;
Δ = b2-4ac
Δ = 222-4·3·(-1)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-4\sqrt{31}}{2*3}=\frac{-22-4\sqrt{31}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+4\sqrt{31}}{2*3}=\frac{-22+4\sqrt{31}}{6}
$