First of all, this function is not one-to-one, so its inverse would
not be a function.  Here is its graph:



To be one-to-one, a function must not only pass the vertical line
test, it also must pass the horizontal line test as well.  As you
can see this graph does not pass the horizontal line test.

However we can restrict the domain in two ways, splitting the
graph into two separate parts, each of which has an inverse.

If we restrict the function to the interval (, ],
we have this graph:

 

To find the inverse



Change  to 



Interchange x and y





Group the 



Use the quadratic formula to solve for y:

 




 




Now if you just take the +, you have



Change the y to 



and the graph of that is:



Now if you put both graphs on the same graph:

 

Now if you plot the identity line , you see that, the inverse
function is the reflection of the original function across the 
identity line 




You can also do the same with the right half of the original graph:



and the equation of that inverse function is

 

Edwin