(4x-3)(5x-5)=0

Solution for (4x-3)(5x-5)=0 equation:

(4x-3)(5x-5)=0

We multiply parentheses ..

(+20x^2-20x-15x+15)=0

We get rid of parentheses

20x^2-20x-15x+15=0

We add all the numbers together, and all the variables

20x^2-35x+15=0

a = 20; b = -35; c = +15;
Δ = b2-4ac
Δ = -352-4·20·15
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5}{2*20}=\frac{30}{40}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5}{2*20}=\frac{40}{40}
=1
$