Remember the definition of continuity - you're trying to show that for any $\varepsilon > 0$ (think of this as very small) you can find some $\delta > 0$ (also very small) so that $|x - 2| < \delta$ implies directly that $|f(x) - f(2)| < \varepsilon$.
You're right to look at $|f(x) - f(2)| = |3x^2 + 2x - 16|$. As Martin Sleziak notes in his comment, this factors as $|3(x-2)(x + 8/3)| = 3|x - 2||x + 8/3|$. Observe that we now have a factor of $|x - 2|$ - this is great news! The only choice we actually get to make in this whole business is how small we're choosing $|x - 2|$. In general, if you're trying to show continuity at $a$, you should be trying to find a factor of $|x - a|$ somewhere, since you have control over this term.
Back to our problem, how do we make $3|x - 2||x + 8/3| < \varepsilon$? We can make $|x - 2|$ as small as we like, so worry about the $|x + 8/3|$ term first. Think of absolute value as distance: if $x$ is very close to 2, then $|x + 8/3|$ is very close to $2 + 8/3 = 14/3$. If you choose $|x - 2| < 1$, then $x < 3$, so $|x + 8/3| < 17/3$. So we have $3|x - 2||x + 8/3| < 3|x - 2|(17/3) = 17|x - 2|$. If $|x - 2| < \varepsilon / 17$, then we have $17|x - 2| < 17(\varepsilon/17) = \varepsilon$ as desired.
Review the choices we made - we need $|x - 2| < 1$ and $|x - 2| < 17 / \varepsilon$. So if you set $\delta = \min(1, 17/\varepsilon)$, then $|x - 2| < \delta$ really does imply $|f(x) - f(2)| < \varepsilon$ by the argument above.