5k^2-26k+24=0

Solution for 5k^2-26k+24=0 equation:

5k^2-26k+24=0

a = 5; b = -26; c = +24;
Δ = b2-4ac
Δ = -262-4·5·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*5}=\frac{12}{10}
=1+1/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*5}=\frac{40}{10}
=4
$