Determine whether $F(x)= 5x+10$ is $O(x^2)$ Please, can someone here help me to understand the Big-O notation in discrete mathematics?

Determine whether $F(x)= 5x+10$ is $O(x^2)$


Short answer:
Yes, it is $\mathcal{O}(x^2)$.
Long answer:
Let's look at the definition of $\mathcal{O}$:

$f(x) = \mathcal{O}(g(x))$ if and only if there exists a positive real number $M$ and a real number $x_0$ such that:
  $$|f(x)| \le M|g(x)| \text{ for all } x \gt x_0$$

In layman's terms, this is saying that $\mathcal{O}(g(x))$ is an upper bound to the function.  Basically, at some point, $g(x)$ grows as fast or faster than $f(x)$.
Clearly, $x^2$ eventually beats out $x$ in terms of growth over the long haul.  In the "long haul," (as $x \to \infty$), it doesn't matter what is added to $x$ or what $x$ is multiplied by--$x^2$ will beat it.  Thus, $x^2$ is an upper bound.
(This is a non-rigorous argument, of course.  Rather than prove it for you, this is just meant to give a feel for how $\mathcal{O}$ works.)
