(2x-4)(5x-3)=0

Solution for (2x-4)(5x-3)=0 equation:

(2x-4)(5x-3)=0

We multiply parentheses ..

(+10x^2-6x-20x+12)=0

We get rid of parentheses

10x^2-6x-20x+12=0

We add all the numbers together, and all the variables

10x^2-26x+12=0

a = 10; b = -26; c = +12;
Δ = b2-4ac
Δ = -262-4·10·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*10}=\frac{12}{20}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*10}=\frac{40}{20}
=2
$