Determine if a set is open 

How can one determine whether the set $$A = \{(x,y)\mid 4\sqrt{3}y-13x^2-7y^2-6\sqrt{3} xy-4x \geq 0\}$$ is open


How can this be done, and what is the actual meaning of a set being open? How can this be illustrated?
 A: The only subsets of $\mathbb R^2$ that are both open and closed are $\mathbb R^2$ itself and $\varnothing.$
The set $\{(x,y)\mid 4\sqrt{3}y-13x^2-7y^2-6\sqrt{3} xy-4x \geq 0\}$ is the inverse-image of the closed set $[0,\infty)$ under the continuous function $(x,y)\mapsto 4\sqrt{3}y-13x^2-7y^2-6\sqrt{3} xy-4x \geq 0.$
That function is continuous because it is a polynomial function.
The inverse-image of a closed set under a continuous function is closed. Therefore the set in question is closed. Therefore it is open only if it is either $\mathbb R^2$ or $\varnothing.$ It is not all of $\mathbb R^2$ since it fails to contain the point $(1,0).$ It is not empty, since, as "Arthur" pointed out, it contains the origin. So it is not open.
A: 
I want to be clear this wasn't really meant to be an answer but only to help you get some intuition and it was a little too long to be a comment

It's closed because it contains the limit points of the set.  Roughly here's the idea:  Suppose you have a sequence of points $(x_n,y_n) \in B:=  \{(x,y)\mid 4\sqrt{3}y-13x^2-7y^2-6\sqrt{3} xy-4x > 0\}\subset A$  (yes I intended to have $>$ and not $\ge$).  Then the limit of $(x_n,y_n)$ may not lie in $B$ but it will lie in $A$..... so $A$ $A$ contains its limit points and is therefore closed.  Almost always when you see $>$ or $<$ think "open" and when you see $\ge$ or $\le$ think "closed".  Of course, you need to actually prove that $A$ contains its limit points.
