What is algebraic geometry?

It's a massive subject, and there are many different perspectives; here are a few that don't require too much background.

Perspective one: It's a generalization of linear algebra.

Linear algebra is about dealing with systems of linear equations. This is easy: the set of solutions to a (homogeneous) system is just some subspace of $F^n$ (where $F$ is the field of scalars), and you can compute its dimension by row-reducing your system into echelon form.

Algebraic geometry is about dealing with systems of polynomial equations. As you may imagine, this is much harder. In linear algebra, much of the theory is entirely independent of the field $F$, at least until you want to diagonalize operators; in algebraic geometry, non-algebraically-closed $F$ are a massive headache, and there are phenomena in characteristic $p$ that don't show up in characteristic $0$.

Perspective two: It's a computational tool in classical geometry.

In geometry and topology we may wish to study invariants of manifolds. We define lots of invariants, e.g., homology groups, but how can we get our hands on them? For most examples, we can't do it easily at all, but if the example happens to be a complex manifold given by polynomial equations, there's a lot more that we can say. This is especially important if you want to do things with computers.

Perspective three: It's a conceptual way to think about commutative algebra.

If I give you some ring, OK, great, it has prime ideals, maximal ideals, zero divisors, etc. What does all this mean, and how do you ever remember the barrage of technical theorems about integrality, Artin rings, regular local rings, etc?

If the ring is the ring of functions on some space, then the geometry of the space may reflect properties of the ring, and we can remember the commutative algebra by picturing the geometry. What Grothendieck realized is that if we define "space" correctly (which is not so easy), every ring is the ring of functions on some space! For an example of how you might relate geometry to intrinsic properties of the ring: the space attached to a ring is connected if and only if all of the zero divisors in the ring are nilpotent.

Suppose you look at all polynomials in two variables, $\mathbb C[x,y]$. Any element $p$ in that ring can be evaluated at any element of $\mathbb C^2$ - if we have two complex values, $a,b$ we can compute $p(a,b)$. And if $p,q$ evaluate to the same value at every point of $\mathbb C^2$ then they are actually the same polynomial.

Now, look at how $\mathbb C[x,y]$ acts when it is evaluated only on some subset of $\mathbb C^2$, say the set of points $X=\{(a,b): b = a^2\}$.

Then $\mathbb C[x,y]$ evaluation is no longer "faithful," but rather, two different polynomials, such as $p_1(x,y)=x^3y$ and $p_2(x,y)=xy^2$, can evaluate as the same on $X$. On X, then, the ring of evaluation functions is actually $R=\mathbb C[x,y]/\left<y-x^2\right>\cong \mathbb C[x]$.

That's a VERY simple case, but Algebraic Geometry is founded on the idea that we can learn something about the geometry of the solutions to a set of equations (in this case, $b=a^2$) by looking at the ring of evaluation functions on that set.

Now, given any point $(a,b)\in X$, there is a natural ring homomorphism $\phi:R\rightarrow \mathbb C$ which corresponds to evaluation. Since this map is onto, this means that the point $(a,b)$ corresponds to some maximal ideal of $R$.

That correspondence between points of the graph of your equations and maximal ideals of the "evaluation ring" associated with your equations, is the big starting point of algebraic geometry.

I will skip the topology part of the question, because it is probably too confusing to try to mix in, except to say that the types of topology used in Algebraic Geometry are very different from the typical starting topologies used in Algebraic Topology.