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

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

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

We multiply parentheses ..

(+8x^2-32x-16x+64)=0

We get rid of parentheses

8x^2-32x-16x+64=0

We add all the numbers together, and all the variables

8x^2-48x+64=0

a = 8; b = -48; c = +64;
Δ = b2-4ac
Δ = -482-4·8·64
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*8}=\frac{32}{16}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*8}=\frac{64}{16}
=4
$