Over number fields, two quadratic forms are equivalent iff they have the same dimension, signature, discriminant and Hasse invariant.


How is the situation like over finitely generated fields?



equivalence of quadratic forms over finitely generated fields
In general, quadratic forms are not determined by their Stiefel-Whitney invariants, see Scharlau, Winfried: Quadratische Formen und Galois-Cohomologie. In: Invent.
Math., 4 (1967), 238–264, p. 251 f.


However, if $\dim{q} \leq 3$, $q$ is determined by dimension, discriminant and Hasse invariant, see Lam, Tsit-Yuen: Introduction to Quadratic Forms Over Fields. Graduate Studies
in Mathematics 67, American Mathematical Society 2005, p. 120, Theorem 3.21.


