(5-x)(4-x)=0

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

(5-x)(4-x)=0

We add all the numbers together, and all the variables

(-1x+5)(-1x+4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-9x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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