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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-8x^2+20x+6x-15)=0

We get rid of parentheses

-8x^2+20x+6x-15=0

We add all the numbers together, and all the variables

-8x^2+26x-15=0

a = -8; b = 26; c = -15;
Δ = b2-4ac
Δ = 262-4·(-8)·(-15)
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*-8}=\frac{-40}{-16}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+14}{2*-8}=\frac{-12}{-16}
=3/4
$