What is the Meaning of $Q[u_1] \cong Q[x]/\langle f(x)\rangle$?

Suppose that $$f(x) = ax^2 + bx + c \in Q[x]$$ is irreducible. Then $$\sqrt{(b^2 − 4ac)} \neq 0$$, so that $$f$$ has two distinct roots -

$$\begin{array}{*{20}c} {u_1 = \frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}}}{, u_2 = \frac{{ - b - \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$$

I found in an article (similar examples can be found in other texts) -

$$Q[u_1] \cong Q[x]/\langle f(x)\rangle.$$

Now I understand -

$$\langle f(x)\rangle$$ is an ideal, $$Q[x]/\langle f(x)\rangle$$ is a quotient ring, elements of $$Q[x]/\langle f(x)\rangle$$ are cosets, i.e. $$Q[x]/\langle f(x)\rangle$$ is the set of polynomials in $$Q[x]$$ of degree less than $$\text{deg}(f(x))$$, i.e. degree $$2$$.

On the other hand, $$Q[u_1]$$ is created by adjoining $$u_1$$, which is an element of degree 2 (the closed expression of $$u_1$$ has square root and square), to $$Q_1$$.

Then how is it possible that $$Q[u_1] \cong Q[x]/\langle f(x)\rangle?$$ What does $$Q[u_1] \cong Q[x]/\langle f(x)\rangle$$ mean? What are the elements of $$Q[x]/\langle f(x)\rangle$$ here? Plz explain elaborately with demonstration. Thanks.

The substitution homomorphism $$\phi(X)\mapsto \phi(u_1)$$ from $$\mathbb{Q}[X]$$ to $$\mathbb{C}$$ has kernel $$\langle f(X)\rangle$$ and image $$\mathbb{Q}[u_1]$$; so the result follows from the Isomorphism Theorem.
Well, put $$u_1= x + \langle f(x)\rangle$$ (the residue class of $$x$$ mod $$f(x)$$).
Then $$Q[u_1]$$ becomes $$Q[x]/\langle f(x)\rangle$$.
It is clear that $$u_1$$ is a zero of $$f(x)$$ in the quotient ring $$Q[u_1]$$.
As a note, if $$f(x)$$ is irreducible the ring adjoint $$Q[u_1]$$ equals the field extension $$Q(u_1)$$.