5k^2-24k+28=0

Solution for 5k^2-24k+28=0 equation:

5k^2-24k+28=0

a = 5; b = -24; c = +28;
Δ = b2-4ac
Δ = -242-4·5·28
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4}{2*5}=\frac{20}{10}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4}{2*5}=\frac{28}{10}
=2+4/5
$