20k^2+29k+5=0

Solution for 20k^2+29k+5=0 equation:

20k^2+29k+5=0

a = 20; b = 29; c = +5;
Δ = b2-4ac
Δ = 292-4·20·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*20}=\frac{-50}{40}
=-1+1/4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*20}=\frac{-8}{40}
=-1/5
$