4k^2+20k+17=0

Solution for 4k^2+20k+17=0 equation:

4k^2+20k+17=0

a = 4; b = 20; c = +17;
Δ = b2-4ac
Δ = 202-4·4·17
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{2}}{2*4}=\frac{-20-8\sqrt{2}}{8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{2}}{2*4}=\frac{-20+8\sqrt{2}}{8}
$