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

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

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

We multiply parentheses ..

(-8x^2-16x+6x+12)=0

We get rid of parentheses

-8x^2-16x+6x+12=0

We add all the numbers together, and all the variables

-8x^2-10x+12=0

a = -8; b = -10; c = +12;
Δ = b2-4ac
Δ = -102-4·(-8)·12
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*-8}=\frac{-12}{-16}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*-8}=\frac{32}{-16}
=-2
$