(x-5)(4x-8)=0

Solution for (x-5)(4x-8)=0 equation:

(x-5)(4x-8)=0

We multiply parentheses ..

(+4x^2-8x-20x+40)=0

We get rid of parentheses

4x^2-8x-20x+40=0

We add all the numbers together, and all the variables

4x^2-28x+40=0

a = 4; b = -28; c = +40;
Δ = b2-4ac
Δ = -282-4·4·40
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12}{2*4}=\frac{40}{8}
=5
$