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

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

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

We multiply parentheses ..

(+6x^2+4x-12x-8)=0

We get rid of parentheses

6x^2+4x-12x-8=0

We add all the numbers together, and all the variables

6x^2-8x-8=0

a = 6; b = -8; c = -8;
Δ = b2-4ac
Δ = -82-4·6·(-8)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*6}=\frac{-8}{12}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*6}=\frac{24}{12}
=2
$