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There are a number of ways (other than using a calculator).

Bracketing

Find two integers, x and y such that x2 < 7 < y2

22 = 4 < 7 < 9 = 32 so sqrt(7) is between 2 and 3.

Next find two integers between 20 and 30 such that x2 < 700 < y2

262 = 676 < 700 < 729 = 272 and so sqrt(7) is between 2.6 and 2.7

Next find two integers between 260 and 270 such that x2 < 70000 < y2

2642 = 69696 < 70000 < 70225 = 2652 and so sqrt(7) is between 2.64 and 2.65

As you can see, you get one additional decimal place at each step. Continue until the desired level of precision is reached.

Newton-Raphson method

This is a more efficient method. If you want to know more about the method, look it up on Wikipedia.

Define f(x) = x2 - 7 so that f(x) = 0 when x = sqrt(7).

Then the derivative of f(x), is f'(x) = 2x

Start with a value of x, say x0 that will make f(x) very approximately 0.

Then calculate x1 = x0 - f(x0)/f'(x0).

Repeat until you have the desired level of accuracy.

Here, for example, if you start with x0 = 2 you will get

x1 = 2 + (4-7)/(2*2) = 2.75

x2 = 2.75 + (2.752-7)/(2*2.75) = 2.6477

x3 = 2.6477 + (2.64772-7)/(2*2.6477) = 2.64575

which is accurate to 5 decimal places.

Incidentally, for finding a square root, a totally outlandish starting point will not matter. Even if you start with 10 as an estimate for sqrt(7), x3 will be accurate to approx one in a thousand; x4 to less than one in a million.

There is also a method resembling long division, but this site is not particularly suited for explaining it.

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Q: How do you figure out the square root of 7?
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