2x(4x-8)=0

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

2x(4x-8)=0

We multiply parentheses

8x^2-16x=0

a = 8; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·8·0
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{-(-16)-16}{2*8}=\frac{0}{16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*8}=\frac{32}{16}
=2
$