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

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

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

We add all the numbers together, and all the variables

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