(8x)(x-1)=0

Solution for (8x)(x-1)=0 equation:

(8x)(x-1)=0

We multiply parentheses

8x^2-8x=0

a = 8; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·8·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*8}=\frac{0}{16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*8}=\frac{16}{16}
=1
$