Yes, your inference is correct. Essentially it is a special case of the well-known discriminant test. Namely, if a quadratic $\rm\:f(x)\in R[x]\:$ has a root in a ring R, then its discriminant is a square in R. Said contrapositively, if the discriminant is not a square in R, then the quadratic has no root in R.
The proof by completing the square works in any ring R (so in $ \mathbb Z/89 = $ integers mod $89$), viz.
$$\rm\: \ \ 4a\:(a\:x^2 + b\:x + c = 0)\:\Rightarrow\: (2a\:x+b)^2 =\: b^2 - 4ac $$
When learning about (modular) arithmetic in new rings it is essential to keep in mind that, like above, any proofs from familiar concrete rings (e.g. $\mathbb Q,\mathbb R,\mathbb C)$ will generalize to every ring if they are purely ring theoretic, i.e. if the proof uses only universal ring properties, i.e. laws that hold true in every ring, e.g. commutative, associative, distributive laws. Thus many familar identities (e.g. Binomial Theorem, difference of squares factorization) are universal, i.e. hold true in every ring.
This is one of the great benefits provided by axiomatization: abstracting the common properties of familiar number systems into the abstract notion of a ring allows one to give universal proofs of ring theorems. It is not necessary to reprove these common ring properties every time one studies a new ring (such reproofs occurred frequently before rings was axiomatized).