Task
Show that $10n^2-15n+2016 = \Omega(n^2)$ by definition
Attempt to show
By definition statement is true if $\exists c, n_0$ so that
$$ 10n^2-15n+ 2016 \ge c n^2 , \text{ when } n \ge n_0$$
Let $c = 1, n_0 = 0$, then we have
$$ 9n^2 - 15n + 2016 \ge 0, \text{ when } n\ge 0$$
which holds because
$$ (-15)^2-4\cdot 2016 < 0 $$
Is this correct?