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

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

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

We multiply parentheses ..

(+4x^2-16x+8x-32)=0

We get rid of parentheses

4x^2-16x+8x-32=0

We add all the numbers together, and all the variables

4x^2-8x-32=0

a = 4; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·4·(-32)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*4}=\frac{32}{8}
=4
$