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

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

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

We multiply parentheses ..

(+20x^2-20x-16x+16)=0

We get rid of parentheses

20x^2-20x-16x+16=0

We add all the numbers together, and all the variables

20x^2-36x+16=0

a = 20; b = -36; c = +16;
Δ = b2-4ac
Δ = -362-4·20·16
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4}{2*20}=\frac{32}{40}
=4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*20}=\frac{40}{40}
=1
$