Your equation is 

, find 

After working on it, I found that there is no solution in integers.

However if the minus sign before the square root on the right side is changed to
plus, there is a solution.  I will prove that I am right. 

I'll first do it with the minus sign, just as was given, show that there is no
solution, then change that minus sign to plus, and find the solution.

First we simplify the left side by writing it as the sum of two square roots of
integers.

, where A and B are positive integers.



Square both sides:

(1)    

Equate the terms with square roots:

 

Divide both sides by 2



Square both sides





Equate the terms of (1) that do not have square roots:



So we have the system of equations:



Solve that by substitution and get {A,B} = {11,16}

So the left side of the original equation is



or



We might stop here to convince ourselves with a calculator
that the left side of the original equation actually does
equal to the above expression.



By calculator, we get the same approximation 7.31662479 for both expressions,
so that satisfies us that the left side of the original equation does simplify
to 

Now the original equation becomes:





Since y is an integer, so is y-4.  

Suppose that integer is A.  Then



Square both sides:



Since x, y, and A are all integers the right side is also an integer.
However the left side is not an integer but is irrational.

Therefore I have proved that your equation as stated has no solution.
That's because an integer cannot equal to an irrational number.

However if the original equation had been this:



Then we would have had this:



The two sides will be exactly the same if y=4 and x-2y=11

Substituting 4 for y in 

x-2y=11
x-2(4)=11
x-8=11
x=19

So the solution with the minus changed to plus is x=19, y=4.

Then, 

You should call your teacher's attention to this.  It has to be an error.

Edwin