We will begin by adding 5 to both sides of the equation. 

Next, let's square both sides to eliminate the radical.

Finally, we can solve this like a simple two-step equation. Subtract 3 from both sides of the equation.

Now, divide each side by 2. 

Finally, check the solution to make sure that it results in a true statement.

When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals.  So a good first step is to add  and subtract  from both sides so that you get:

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

Then when you subtract  from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

This factors to:

Which would seem to yield solutions of  and .  However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions.  Here if you plug in  the math holds:  because , and .

But if you plug in , you'll see that the original equation is not satisfied:  because  does not equal .  Therefore the only proper answer is .

To solve, first square both sides:

squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply :

This is a quadratic, we just need to combine like terms and get it equal to 0

now we can solve using the quadratic formula:

This gives us 2 potential answers:

and

Since the expression under the radical cannot be negative,

.

Solve for x:

This is the domain, or possible  values, for the function.

Because the term  inside the radical in the numerator has to be greater than or equal to zero, there is a restriction on our domain; to describe this restriction we solve the inequality for :

Add  to both sides, and the resulting inequality is:

     or   

Next, we consider the denominator, a quadratic; we know that we cannot divide by zero, so we must find x-values that make the denominator equal zero, and exclude them from our domain:

We factor an  out of both  and  :

Finding the zeroes of our expression leaves us with:

Therefore, there are 3 restrictions on the domain of the function:

The graph of  is: 

This graph is translated 2 units left and 4 units down from the graph of , which looks like:

The number 4 and its associated negative sign, which fall outside of the square root, tell us that we need to shift the graph down four units. If you take what's under the square root, set it equal to 0, then solve, you'll get x=-2, which you can interpret as being shifted left two units. 

Each of the graphs below that are incorrect either improperly shift the graph right (instead of left), up (instead of down), or both right and up. 

The above graph is not a function. In order to be a function, the graph must pass the vertical line test. If you look at x=2, you can see that x=2 is a value on both the green graph and the purple graph. Therefore it fails to pass the vertical line test and is not a function.

This graph is a function. We call this type of function a piecewise (or step) function. This type of function is defined in a certain way for certain x values, and then in a different way for different x values. However, it still passes the vertical line test, so it is a function!

Firstly, this is a function as it passes the vertical line test.

To find the domain, begin at the far left of the graph, starting with the red portion. You can trace along it continuously until you get to the point (-5,-5). At this point, it jumps upward to the green graph. While the y values jump, there is no gap in the x values. You can trace from left to right along the green graph until you get to x=2. Then, the graph jumps up to the purple values. Once again, there is no gap in the x values. The purple graph is continuous and goes onwards to positive infinity. Therefore, the domain of this graph is all real numbers or .

Next, let's start again at the bottom of the graph (in red), this time looking at the y values. We begin at negative infinity and can trace continuously until we get to the point (-5,-5). There is a discontinuity between the red and green graphs' y values. We use the  symbol ("union") to connect the disconnected parts of the graph. The green graph picks up at -1.71 and the y values continue upward until they get to 1.256. Once again, there is a discontinuity in the range, so we again use the union symbol. The y values pick back up at the point (2,4) and continue upwards to infinity. Therefore the range of this function is .

The graph of  is

Keep in mind that you cannot put a negative number under a square root and get a real number answer. If we tried to plug in -2 for x, it would be undefined. That is why the graph only shows positive values of x. 

The other graphs pictured below are , , and .