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

Solution for (2x-4)(6x+8)=0 equation:

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

We multiply parentheses ..

(+12x^2+16x-24x-32)=0

We get rid of parentheses

12x^2+16x-24x-32=0

We add all the numbers together, and all the variables

12x^2-8x-32=0

a = 12; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·12·(-32)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-40}{2*12}=\frac{-32}{24}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+40}{2*12}=\frac{48}{24}
=2
$