2k^2-5k-17=0

Solution for 2k^2-5k-17=0 equation:

2k^2-5k-17=0

a = 2; b = -5; c = -17;
Δ = b2-4ac
Δ = -52-4·2·(-17)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{161}}{2*2}=\frac{5-\sqrt{161}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{161}}{2*2}=\frac{5+\sqrt{161}}{4}
$