Find x such that x - 4, x, and 3x - 8 are three consecutive terms in a geometric sequence

In a geometric sequence dividing any term by the term before it yields the common multiplier.
Then x/(x+8) = (3x-8)/x
Solving for x:  reduces to x^2-10x +16 factor into (x-2)(x-8)
Then x = 2 or 8
2 produces a sequence of -2, 2, -2 ...
8 produces a sequence of 4, 8, 16 ...

2x2 =4 welcome