The matrix $H$ is the Hessian matrix of second partial derivatives. We find from "congruence diagonalization" that the eigenvalues of $H$ are $++- \; \; .$ Of course, the diagonal entries of $D$ are not the eigenvalues themselves. This is about Sylvester's Law of Inertia.
Sylvester's law of inertia is a theorem in matrix algebra about
certain properties of the coefficient matrix of a real quadratic form
that remain invariant under a change of basis. Namely, if $A$ is the
symmetric matrix that defines the quadratic form, and $S$ is any
invertible matrix such that $D = SAS^T$ is diagonal, then the number
of negative elements in the diagonal of $D$ is always the same, for
all such $S \; ; \; $ and the same goes for the number of positive
elements.
If I had the energy to find the eigenvectors, I could write $O^T H O = D_0$ with orthogonal $O$ and diagonal $D_0,$ at which point the diagonal entries of $D_0$ would be the eigenvalues. Sylvester says that my $ P^T H P = D $ gives the same number of positive diagonal entries and the same number of negative, also the same number of zero entries. For this example, two positive eigenvalues and one negative.
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 4 }{ 3 } & 1 & 0 \\
- \frac{ 27 }{ 10 } & \frac{ 7 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
6 & 8 & 5 \\
8 & 14 & 2 \\
5 & 2 & 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & - \frac{ 27 }{ 10 } \\
0 & 1 & \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & \frac{ 10 }{ 3 } & 0 \\
0 & 0 & - \frac{ 67 }{ 10 } \\
\end{array}
\right)
$$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = \left(
\begin{array}{rrr}
6 & 8 & 5 \\
8 & 14 & 2 \\
5 & 2 & 4 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
1 & \frac{ 4 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
6 & 0 & 5 \\
0 & \frac{ 10 }{ 3 } & - \frac{ 14 }{ 3 } \\
5 & - \frac{ 14 }{ 3 } & 4 \\
\end{array}
\right)
$$
==============================================
$$ E_{2} = \left(
\begin{array}{rrr}
1 & 0 & - \frac{ 5 }{ 6 } \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{2} = \left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & - \frac{ 5 }{ 6 } \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rrr}
1 & \frac{ 4 }{ 3 } & \frac{ 5 }{ 6 } \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & \frac{ 10 }{ 3 } & - \frac{ 14 }{ 3 } \\
0 & - \frac{ 14 }{ 3 } & - \frac{ 1 }{ 6 } \\
\end{array}
\right)
$$
==============================================
$$ E_{3} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{3} = \left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & - \frac{ 27 }{ 10 } \\
0 & 1 & \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{3} = \left(
\begin{array}{rrr}
1 & \frac{ 4 }{ 3 } & \frac{ 5 }{ 6 } \\
0 & 1 & - \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{3} = \left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & \frac{ 10 }{ 3 } & 0 \\
0 & 0 & - \frac{ 67 }{ 10 } \\
\end{array}
\right)
$$
==============================================
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 4 }{ 3 } & 1 & 0 \\
- \frac{ 27 }{ 10 } & \frac{ 7 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
6 & 8 & 5 \\
8 & 14 & 2 \\
5 & 2 & 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 4 }{ 3 } & - \frac{ 27 }{ 10 } \\
0 & 1 & \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & \frac{ 10 }{ 3 } & 0 \\
0 & 0 & - \frac{ 67 }{ 10 } \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 4 }{ 3 } & 1 & 0 \\
\frac{ 5 }{ 6 } & - \frac{ 7 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & \frac{ 10 }{ 3 } & 0 \\
0 & 0 & - \frac{ 67 }{ 10 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 4 }{ 3 } & \frac{ 5 }{ 6 } \\
0 & 1 & - \frac{ 7 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
6 & 8 & 5 \\
8 & 14 & 2 \\
5 & 2 & 4 \\
\end{array}
\right)
$$