How do you find the square root of 13?

There is no simpler form for 13's square root because it is a prime number, which is the explanation. ${\displaystyle \sqrt{13}}$ is an irrational number somewhere between ${\displaystyle 3=\sqrt{9}}$ and ${\displaystyle 4=\sqrt{16}}$
A good first approximation using linear interpolation would be:
${\displaystyle \sqrt{13}\approx 3.6=\frac{18}{5}}$
From our initial guess, we can obtain more accurate ones. (call it ${\displaystyle a_0}$) utilizing a Newton-Raphson approach.
Typical formula for obtaining a closer approximation to ${\displaystyle \sqrt{n}}$ would be:
${\displaystyle a_{i+1}=\frac{a_i^2+n}{2a_i}}$
I prefer to separate ${\displaystyle a_i}$ into numerator ${\displaystyle p_i}$ and denominator ${\displaystyle q_i}$. So ${\displaystyle a_i=\frac{p_i}{q_i}}$ and we can iterate using the formula:
${\displaystyle \{\begin{array}{c}{p}_{i+1}=p_i^2+n{q}_i^2\\ q_{i+1}=2p_i{q}_i\end{array} }$
In our example ${\displaystyle n=13,p_0=18,q_0=5}$ and we find:
${\displaystyle \{\begin{array}{c}{p}_1=p_0^2+13q_0^2=324+13\cdot 25=649\\ q_1=2p_0{q}_0=180\end{array} }$
Our approximate figure if we stopped here would be:
${\displaystyle \sqrt{13}\approx \frac{649}{180}=3.60\bar{5}}$
Let's try one more iteration:
${\displaystyle \{\begin{array}{c}{p}_2=p_0^2+13q_1^2=421+13\cdot 32400=842401\\ q_2=2p_1{q}_1=233640\end{array} }$
Stopping here, we have:
${\displaystyle \sqrt{13}\approx \frac{842401}{233640}\approx 3.60555127547}$
Using a calculator:
$\sqrt{13}\approx 3.60555127546398929311$

Explanation: 
Look for a generalised continued fraction of the form: 
${\displaystyle \sqrt{13}=a+\frac{b}{2a+\frac{b}{2a+\frac{b}{2a+\dots }}}}$ 
${\displaystyle =a+\frac{b}{a+\sqrt{13}}}$ 
Multiply both ends by ${\displaystyle (a+\sqrt{13})}$ to find: 
${\displaystyle a\sqrt{13}+13=a^2+a\sqrt{13}+b}$ 
Hence: 
${\displaystyle b=13-a^2}$ 
Choose a reasonable approximation that is somewhat smaller than to ensure that our generalized continuous fraction converges rapidly ${\displaystyle \sqrt{13}}$. 
Note that ${\displaystyle 3=\sqrt{9}<\sqrt{13}<\sqrt{16}=4}$ In order to find a good approximation, we use linear interpolation: 
${\displaystyle \sqrt{13}\approx 3.6=\frac{18}{5}}$ 
Note also that: 
${\displaystyle {\left(\frac{18}{5}\right)}^2=\frac{324}{25}<\frac{325}{25}=13}$ 
So let ${\displaystyle a=\frac{18}{5}}$ and ${\displaystyle b=13-a^2=\frac{1}{25}}$ to get: 
${\displaystyle \sqrt{13}=\frac{18}{5}+\frac{\frac{1}{25}}{\frac{36}{5}+\frac{\frac{1}{25}}{\frac{36}{5}}+\dots }}$ 
To obtain an accurate estimate that is rational, we can truncate this continuous fraction.
For instance:
${\displaystyle \sqrt{13}\approx \frac{18}{5}+\frac{\frac{1}{25}}{\frac{36}{5}}=\frac{18}{5}+\frac{1}{180}=\frac{649}{180}=3.60\bar{5}}$ 
Or: 
${\displaystyle \sqrt{13}\approx \frac{18}{5}+\frac{\frac{1}{25}}{\frac{36}{5}+\frac{\frac{1}{25}}{\frac{36}{5}}}=\frac{23382}{6485}\approx 3.60555127}$