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

Solution for (4x-1)(5x-1)=0 equation:

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

We multiply parentheses ..

(+20x^2-4x-5x+1)=0

We get rid of parentheses

20x^2-4x-5x+1=0

We add all the numbers together, and all the variables

20x^2-9x+1=0

a = 20; b = -9; c = +1;
Δ = b2-4ac
Δ = -92-4·20·1
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*20}=\frac{8}{40}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*20}=\frac{10}{40}
=1/4
$