For which $a$ is the matrix negative definite? Find the number of integer numbers of $a$ such that the matrix $$\begin{pmatrix}-1& 0 & 5 \\ 0 & a-50 & 0 \\ 5 & 0 & 6-a\end{pmatrix}$$ is negative definite.
For that, the eigenvalues must be negative, right?
I calculated the eigenvalues using Wolfram and we get these.
So all these have to be negative and we get a system of inequalities :
$$a-50<0 \\ \frac{1}{2}\left (-\sqrt{a^2-14a+149}-a+5\right ) <0 \\ \frac{1}{2}\left (\sqrt{a^2-14a+149}-a+5 \right )<0$$ So we get from the first $a<50$, from the second $a\in \mathbb{R}$ and from the last one $a>31$.
So the integer values for $a$ are from $32$ to $49$, so there are $49-32+1=18$ values, right?

Your approach is correct, but it could be slightly simpler: with the help of the rule of Sarrus or by other means, it's easy to find the characteristic polynomial is:
$$(\lambda-a+50)(\lambda^2+(a-5)\lambda+a-31)$$
The eigenvalues must all be negative, so, for the first factor, that means $a<50$. For the second factor, instead of computing the roots, note that the sum of the roots must be negative, while the product of the roots must be positive. With Vieta's formulas, that means $a>31$ and $a>5$. So the condition is indeed $31<a<50$.
