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

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

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

We multiply parentheses ..

(+20x^2-15x-32x+24)=0

We get rid of parentheses

20x^2-15x-32x+24=0

We add all the numbers together, and all the variables

20x^2-47x+24=0

a = 20; b = -47; c = +24;
Δ = b2-4ac
Δ = -472-4·20·24
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-17}{2*20}=\frac{30}{40}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+17}{2*20}=\frac{64}{40}
=1+3/5
$