[SOLVED] Inequality. If x is real, why does it mean this equation has real roots?

Hiya. There's an example in a book that I can follow the algebra of, but I don't quite understand the reasoning of one of the steps. The underlined part is the *one and only* part I don't understand. It would be great if someone could explain it some more.

Here's what the book has:

Q: If x is real, find the possible values of the function: $\displaystyle \frac{x^2}{x+1}$

A: Use $\displaystyle y = \frac{x^2}{x+1} \Rightarrow x^2-yx-y=0 $

*Since x is real, the roots of this equation are real,* so $\displaystyle b^2-4ac \ge 0$

Then there's a bit of algebra to get: $\displaystyle y\le-4$ or $\displaystyle y\ge0$

The algebra's fine and I get that if the roots are real, then $\displaystyle b^2-4ac \ge 0$, but I don't see why x being real means that that equation must have real roots.

If anyone could help me get some more understanding of this, intuitive or otherwise, it'd be greatly appreciated. Thanks.