(r-4)(r-5)=0

Solution for (r-4)(r-5)=0 equation:

(r-4)(r-5)=0

We multiply parentheses ..

(+r^2-5r-4r+20)=0

We get rid of parentheses

r^2-5r-4r+20=0

We add all the numbers together, and all the variables

r^2-9r+20=0

a = 1; b = -9; c = +20;
Δ = b2-4ac
Δ = -92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*1}=\frac{10}{2}
=5
$