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

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

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

We multiply parentheses ..

(+2x^2+4x-8x-16)=0

We get rid of parentheses

2x^2+4x-8x-16=0

We add all the numbers together, and all the variables

2x^2-4x-16=0

a = 2; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·2·(-16)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*2}=\frac{16}{4}
=4
$