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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+2x^2+16x-4x-32))=0


We calculate terms in parentheses: -((+2x^2+16x-4x-32)), so:

(+2x^2+16x-4x-32)

We get rid of parentheses

2x^2+16x-4x-32

We add all the numbers together, and all the variables

2x^2+12x-32

Back to the equation:

-(2x^2+12x-32)


We get rid of parentheses

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

a = -2; b = -12; c = +32;
Δ = b2-4ac
Δ = -122-4·(-2)·32
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*-2}=\frac{32}{-4}
=-8
$