There is a theorem that states that two quadratic forms over $\mathbb{Q}_p$ are equivalent iff they have the same rank, discriminant and the same $\epsilon$ invariant.
(The last is defined as follows: if $f = \sum_{i=1}^n a_ix_i^2$ then $\epsilon(f) = \prod_{1\le i < j \le n} (a_i,a_j)$, where $(\cdot,\cdot)$ is the Hilbert symbol over $\mathbb{Q}_p$.)
The proof (from Serre's Arithmetic) is based on a previous theorem, which relates these invariants to the elements the forms represent.
However, I do not know how I can show that two particular forms are equivalent (with the same invariants of course). So fix some prime $p$ and consider the following two forms over $\mathbb{Q}_p$:
\begin{equation} f(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2 - (x_5^2 + x_6^2 + x_7^2) \nonumber \end{equation} and \begin{equation}
g(x) = -(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2) \nonumber \end{equation}
How can I show in particular that these two are equivalent (without using the classification theorem above)? Thanks.