Solving a recurrence essentially means getting rid of the recursion and giving a way to calculate the answer such a recursive function would give you, without doing the recursion. Additionally, when you solve it in terms of asymptotic complexity, then it becomes a bit easier, since you don't care about the slower-growing terms and can get rid of a lot of terms on the way.
So, choosing an example, if we have, $T(n) = T\left(\frac{n}{2}\right) + 18$ then $T(n) = \Theta\left(\log n\right)$. See, I got rid of the recursion; there is no longer any recursion on the right-hand-side. How did I do this? Well, there are two ways.
General way
We recurse a few steps, and generalize it to step $k$:
$$
\begin{eqnarray}
T(n)&=&T\left(\frac{n}{2}\right) + 18\\
T\left(\frac{n}{2}\right)&=&T\left(\frac{n}{4}\right) + 18\\
T\left(\frac{n}{4}\right)&=&T\left(\frac{n}{8}\right) + 18\\
&...&\\
T\left(\frac{n}{2^k}\right)&=&T\left(\frac{n}{2^{k+1}}\right) + 18\\
T\left(1\right)&=&\Theta(1)&\genfrac{}{}{0}{}{\text{We choose a (constant) base}}{\text{ case, unless given one}}\\
\end{eqnarray}
$$
Then we see how large $k$ must be in order to get down to $T(1)=T\left(\frac n n\right)$:
$$
\begin{eqnarray}
2^{\max k}&=&n\\
{\max k} &=& \Theta \left(\log n\right)\\
\end{eqnarray}
$$
Now we know it will recurse $\max k$ times, so we replace each $T(x)$ on the right-hand-side with its recursed value (calculated above), then generalize it to $\max k$.
$$
\begin{eqnarray}
T(n)&=&T\left(\frac{n}{2}\right) + 18\\
&=&\left(T\left(\frac{n}{4}\right) + 18\right) + 18
& \text{Replace }T\left(\frac{n}{2}\right)\text{ in terms of }T\left(\frac{n}{4}\right)\\
&=&\left(\left(T\left(\frac{n}{8}\right) + 18\right) + 18\right) + 18& ...\\
&...&\\
&=&T\left(\frac{n}{2^{\max k}}\right) + 18\cdot {\max k}\\
&=&T\left(1\right) + 18\cdot \Theta \left(\log n \right)\\
&=&\Theta \left(\log n \right)\\
\end{eqnarray}
$$
Master theorem
An easier way for special cases of asymptotic recurrence relations is called the master theorem, and allows you to "cheat" if $T(n)$ follows 3 specific forms, and you want the asymptotic solution. See the master theorem page for the different forms, but I will demonstrate for the example I chose:
General form:
$T(n) = a \; T\!\left(\frac{n}{b}\right) + f(n) \;\;\;\; \mbox{where} \;\; a \geq 1 \mbox{, } b > 1$
So for us, $a=1,b=2,f(n)=18$. Therefore, $\log_b a = 0$ This fits Case 2 as follows:
Case 2:
If:
$f(n) = \Theta\left( n^{c} \log^{k} n \right)$, where $c = \log_b a$, and some $k \ge 0$.
Then:
$T(n) = \Theta\left( n^{c} \log^{k+1} n \right)$
We have $c=\log_b a=0$ and choose $k=0$. This gives us:
$$\begin{eqnarray}
f(n) &\stackrel{?}{=}& \Theta\left( n^{c} \log^{k} n \right)\\
&\stackrel{?}{=}& \Theta\left( n^{0} \log^{0} n \right)\\
&\stackrel{?}{=}& \Theta\left( 1 \right)&\checkmark\\
\end{eqnarray}$$
Which is true, so we match the case. Therefore:
$$\begin{eqnarray}
T(n) &=& \Theta\left( n^{c} \log^{k+1} n \right)\\
&=& \Theta\left( n^{0} \log^{0+1} n \right)\\
&=& \Theta\left( \log n \right)\\
\end{eqnarray}$$
