5k^2-27k+28=0

Solution for 5k^2-27k+28=0 equation:

5k^2-27k+28=0

a = 5; b = -27; c = +28;
Δ = b2-4ac
Δ = -272-4·5·28
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-13}{2*5}=\frac{14}{10}
=1+2/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+13}{2*5}=\frac{40}{10}
=4
$