How many values of the integer k will make the triangle with sides 6, 8 and k obtuse?
6,8,10 is a right triangle therefore if we enlarge the hyp, then the right angle will become obtuse?
Isn't it only 10 < k <14? How do I get the others using a systematic method?
$k$ has to be in $(2,14)$ for the triangle to exist.
Let's note $\hat{A}$, $\hat{B}$ and $\hat{C}$ the angles between the sides of length respectively $6$ and $8$, $6$ and $k$ and $k$ and $8$.
The side opposing the obtuse angle in an obtuse triangle is the longest of the three and $6<8$, so $\hat{C}$ can never be obtuse.
As you noticed, $\hat{A}$ is obtuse iff $k^2>6^2+8^2$, i.e. $k>10$.
By the same reasonning, $\hat{B}$ is obtuse iff $8^2>6^2+k^2$, i.e. $k<\sqrt{28}=2\sqrt{7}$
Finally, the triangle is obtuse iff $k \in (2;2\sqrt{7})\cup(10;14)$