If $m$ is an odd integer and $m$ and $a$ are relatively prime, we can multiply the inverse of $a$ and complete the square to rewrite $$ax^2 + bx + c ≡ 0 \mod m $$ as $$y^2 ≡ d \mod m$$
express $d$ in terms of $a$, $b$, $c$
find x in terms of $a$, $b$, $c$, and $y$
find respective conditions on d so the original congruence has 0, 1, or 2 solutions.
I'm mainly just confused by what this means to multiply the inverse of a. Is it just saying to multiply by 1/a? Past that, completing the square would allow us to have a term $(a + b)^2$ which would allow us to just say $y = a + b$ creating the $y^2$ term in the second congruence. How does the $d$ come in to the second congruence and what is the relevance of $a$ and $m$ being relatively prime?
The usual construction is to multiply by $4a$ to get
$$4a^2+4ab+4ac \equiv 0 \pmod(m).$$
Then adding and substracting $b^2$ gives
$$4a^2+4ab +b^2 -(b^2-4ac)\equiv (2ax+b)^2-(b^2-4ac)\equiv 0 \pmod{m}.$$
Let $y=2ax+b$ and $d=b^2=4ac.$ Then the solution depends on the quadratic character of $d$. You need the inverse of $a \pmod{m}$ when you solve for $x$ once you know $y$.