I am hoping you understand my solution here because I have really simplified what others are saying above using different concepts, hopefully a simpler approach:
The parametric equation is x = t^2 - 2...
If you draw that graph out, it is x = t^2 and then translated 2 downwards. Think of it was y = x^2 but with the variables replaced and y = f(x) so in this case x = f(t). If you do x = f(t) - 2 you will get x = t^2 - 2 which is a translation of f(t) or t^2 by 2 units downwards (down the y/vertical-axis).
What you've got to realise is that x = t^2 graph is symmetrical because b^2 = 4ac , it has two equal roots of 0.
Therefore, when you translate it 2 units downwards - it is still symmetrical. So, if when t > 0, we have acknowledged that t^2 - 2 > 0, then when t < 0, t^2 - 2 will still be > 0 because the graph is symmetrical - it is a reflection in the line t = 0 or the x-axis. I hope that makes things clearer, it is a symmetrical graph and therefore x > -2 when -∞ < t < ∞, t ≠ 0. This is the same as t > 0 or t < 0 lol.