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) $
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$ 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) $
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$ 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) $
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$ 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) $
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$ 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) $