Let us see how you get the answer without using derivatives. What is the definition of continuity?
A function $f(x)$ is said to be continuous at $x=a$ if for all $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-a|< \delta$ implies that $|f(x) - f(a)| < \epsilon$.
So let $\epsilon > 0$ be given as in your problem ($\epsilon = 0.003$).
Now if $|x + 3| < \delta$, then we see that
$-\delta < x+3 < \delta
\implies -4\delta - 23 < 4x - 11 < 4\delta - 23. $
The question is, "Now that we have a bound on $4x - 11$ of the form $a < 4x - 11 < b$, how can I make it into the form $-c < 4x - 11 <c$?" $a,b$ and $c$ are just some real numbers.
The reason I would like to this is in order for me to put a bound on the absolute value of $4x - 11$.
Notice that $4\delta - 23 < 4\delta + 23$, so that
$-(4\delta + 23) < 4x - 11 < 4\delta - 23 < 4\delta + 23$.
In other words you can be confident now that
$|4x - 11| < 4\delta + 23$.
So if you consider $|f(x) - 36|$, you see that $|x +3| < \delta$ will lead to the conclusion that
$|f(x) - 36|= |(x+3)(4x - 11)| < \delta(4\delta + 23)$.
If you solve $\delta(4\delta + 23) = \epsilon = 0.003$ for delta, you will get
$\delta = 0.0001304$ or $\delta = -5.7501304$. Since delta is positive, you will only have one choice of delta, which is 0.0001304 and you're done!
No guessing, playing around with numbers, using derivatives and working blindly.