(5x-3)(4x-8)=0

Solution for (5x-3)(4x-8)=0 equation:

(5x-3)(4x-8)=0

We multiply parentheses ..

(+20x^2-40x-12x+24)=0

We get rid of parentheses

20x^2-40x-12x+24=0

We add all the numbers together, and all the variables

20x^2-52x+24=0

a = 20; b = -52; c = +24;
Δ = b2-4ac
Δ = -522-4·20·24
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-28}{2*20}=\frac{24}{40}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+28}{2*20}=\frac{80}{40}
=2
$