(k-1)(k-5)=0

Solution for (k-1)(k-5)=0 equation:

(k-1)(k-5)=0

We multiply parentheses ..

(+k^2-5k-1k+5)=0

We get rid of parentheses

k^2-5k-1k+5=0

We add all the numbers together, and all the variables

k^2-6k+5=0

a = 1; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·1·5
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{-(-6)-4}{2*1}=\frac{2}{2}
=1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*1}=\frac{10}{2}
=5
$