How many obtuse-angled triangles have longest side $19$ and other sides $8$ and $x$, where $x$ is an integer? 
In an obtuse angled triangle, the greatest side is $19$ cm and the other two sides are 8 and x cms respectively. If x can only be an integer, then how many such triangles are possible?

I am attaching my way of thinking and solution also here. Please check and give a reply.
However, the real confusion lies in two different theories which are in my mind.
( 1 ) In a triangle, the sum of any two sides of the triangle has to be greater 
      than the third side. So, that gives us the fact that x has to be greater 
      than 11.
( 2 ) Again, in a triangle, the difference between any two sides of the triangle 
      has to be less than the third side and so, it gives the values of x upto 
      19 from 11.
( 3 ) Now, I just read another property somewhere that the sum of squares of the 
      two sides of the triangle has to be less than the square of the third 
      side. So, I dont know whether this property is right or wrong but 
      according to this, the values of x can remain only upto 17 and not 19.
So, a dilemma has been created by the use of the things mentioned in the points
2 and 3.
What to do ?
 A: x has to be greater than 11 for a triangle to be formed (so that $x+8>19$).  By Pythagoras theorem, if the triangle were a right triangle, x would be  17.23
So any integral $x>17$ would mean obtuseness is lost.  Also x cannot be greater than 19 anyway since the longest side is 19.  
Not considering the trivial triangle with x=11, so, x can be 12,13,14,15,16 or 17.
A: Consider the law of cosines: $a^2=b^2+c^2-2bc\cos A$. So for your triangle let's substitute all the values we know: $$19^2=8^2+x^2-2(8)(x)\cos A.$$ It is given as an obtuse triangle, so $\cos A < 0$ (and $> -1$ as always). Lets make the equation above into a quadratic with $0$ on one side: $$x^2-16\cos A x -297=0.$$ Consider when $\cos A = -1$: the positive solution of it is $x=11$. Now consider when $\cos A = 0$: the positive solution is $x=\sqrt{297}$. Since the last side can only be an integer, consider the integer values strictly in the range $(11,\sqrt{297})$. That leaves $6$ possible integer values.
A: 
$$\left\{\begin{array}{c}
x^2+y^2 = n^2 \\
(19-x)^2+y^2 = 64
\end{array}
\right\}
\implies
\left\{\begin{array}{c}
x = \dfrac{n^2+297}{38} \\
y = \dfrac{\sqrt{(27^2-n^2)(n^2-11^2)}}{38}
\end{array} \right\}$$
