Simplest way to balance any chemical equation
You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.

Your chemical equation contains 3 atomic species: $\ce{C}$, $\ce{H}$ and $\ce{O}$. This means that you need 3 equations to balance. First I write the chemical reaction as follows:
$$\ce{a C12H26 + b O2 -&gt; c CO2 + d H2O}$$ 
Now I will write the 3 equalities for the 3 atomic species that we have:
\begin{align}
12a &amp;= c \tag{for C}\\
26a &amp;= 2d \tag{for H}\\
2b  &amp;= 2c + d \tag{for O}
\end{align}
The numerical constants in front of $a$, $b$, $c$ and $d$ come from the number of atoms that are in the molecule for which they are included. 

As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns ($a$ to $d$) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,1 just against convention. This means that we can simply set one of the unknowns to any value. Let's say $a = 1$ for now and revisit this choice after solving the equations.

With $a = 1$ it immediately follows that $c = 12$ and $d = 13$. From those two it then follows that $b = 18.5$. Our balanced reaction becomes:2
$$\ce{C12H26 + 18.5 O2 -&gt; 12 CO2 + 13 H2O}$$ 

In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of $a = 1$ and pick it such that all numbers will become integers. In this case you can easily see that this will happen for $a = 2$ which then results in 
$$\ce{2C12H26 + 37 O2 -&gt; 24 CO2 + 26 H2O}$$

If for some reason it is not easy to see which value for $a$ you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the greatest common divisor of the 4 numbers and divide by that to obtain the same equation.



Notes


Compare $10=10$ to the multiplied version $30=30$, both are correct, but they are just different by a factor $3$.
I'm not sure whether you are familiar with linear algebra, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.