How can I find the square root using pen and paper?

You can use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to arrive at a "long-division" like method. Let me show you how it is done for 3838 before giving the algorithm.

1. Start with 3838
2. Write the digits in groups of 2, so write 3838 as 38,38.
3. For the "highest placed group of 2", find the biggest square number less than it. In this case, $6^2 = 36 < 38 < 49 = 7^2$. So you write down 6, and subtract 36 from the first group. You then get
4. 2,38 = 38,38 - 36,00.
5. Take the 6 you wrote down before, multiply it by 20 (so you get 120). Now find the largest multiple of $6\times 20$ that is less than 238. You'll see that it is 1. So you write down 1 (so your number is now 61). Subtract from 238 120 to get 118.
6. Subtract from 118 the number $1 = 1^2$ to get 117. Now add two more digits to it (to form 117.00). Take 61, multiply it by 20, you get 1220. Find the largest multiple of 1220 less than 11700, which would be $1220 \times 9 = 10980$. So write down 9, and your number is 61.9, and subtract from 11700 the number 10980 to get 720. Subtract from 720 the number $9^2 = 81$ to get 639.

So you've arrived at, at this point, $3838 = (61.9)^2 + 6.39$. And you can continue the process indefinitely.

How does this work? Given a number $A$, you want to find its square root in base-10 representation. Suppose your square root looks like $$ a_{100}a_{10}a_1.a_{0.1}a_{0.0.1}\ldots $$ when expanded as a string of digits. Then you find the biggest $a_{100}$ such that $$ (a_{100}\times 100)^2 \leq A $$ (similar to how you do long division). Then to make solve for the next digit, you use that $$ (a_{100} \times 100 + a_{10}\times 10)^2 \leq A $$ (since you are truncating the decimal expansion, which makes the number smaller). So you solve for the best $a_{10}$ such that $$ (a_{100})^2\times 100^2 + \left[ 20 \times (a_{100}\times a_{10}) + (a_{10})^2\right] \times 100 + \leq A $$ The above expression shows why in the first step you want to group the digits in twos: in some sense we are thinking of $A$ as in base-100, the "square" of base-10.

At every step you use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to compute the next digit correction to the square root.

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Another method, as the Babylonians did it, was recently detailed by John Baez at his blog, which uses the "equality case" of the arithmetic-mean-geometric-mean inequality to power the iteration.

The best method I know of is the recursive series:$$x_1=b\;\;\;\;\;\;\;\;\;x_{n+1}=\frac{1}{2}\left(x_n+\frac{b}{x_n}\right)$$ It converges very rapidly to $\sqrt{b}$ - for example for b = 3, it is accurate to 7 decimal places after only 4 terms. The long division might not be very easy to carry out strictly with pencil and paper, but it is doable.

If your number is close enough to a perfect square, you can use the expansion $$ \sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 +- \cdots + (-1)^{n-1}\frac{(2n-3)!!}{2\cdot(2n)!!}x^n+\cdots $$ where $$ (2n)!! = (2n)(2n-2)\cdots4\cdot2,\quad(2n+1)!!=(2n+1)(2n-1)\cdots3\cdot1. $$

A famous physist R. Feynman is said to have used this formula and beat an abacus.