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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

32x^2-16x-16=0

a = 32; b = -16; c = -16;
Δ = b2-4ac
Δ = -162-4·32·(-16)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-48}{2*32}=\frac{-32}{64}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+48}{2*32}=\frac{64}{64}
=1
$