For any odd prime $p$, a quadratic is solvable mod $p^2$ if it is solvable mod $p$, and $p$ does not divide the discriminant and leading coefficient.

Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p\nmid a$ and $p\nmid D$, where $D=b^2-4ac$. Let $x$ be an integer such that $$ax^2+bx+c\equiv0\pmod{p},\tag{1}$$ so $ax^2+bx+c=mp$ for some integer $m$.

Exercise: Show that $p$ does not divide $2ax+b$, because $p\nmid a$ and $p\nmid D$ and $p$ is odd.

If $2ax+b\equiv0\pmod{b}$ then because $p$ is odd and $p\nmid a$ we have $$x\equiv-b(2a)^{-1}\pmod{p}.$$ Plugging this into $(1)$ yields \begin{eqnarray*} 0&\equiv&a(-b(2a)^{-1})^2+b(-b(2a)^{-1})+c\\ &\equiv&(2a)^{-2}(ab^2-2ab^2+4a^2c)\\ &\equiv&(2a)^{-2}(-aD), \end{eqnarray*} contradicting the fact that $p\nmid a$ and $p\nmid D$ and $p$ is odd.

Now if $y$ is an integer such that $$ay^2+by+c\equiv0\pmod{p}^2,\tag{2}$$ then reducing mod $p$ shows that $y$ is a solution to $(1)$. So it makes sense to suppose that $y=x+kp$ for some integer $k$. Then \begin{eqnarray*} ay^2+by+c&=&a(x+kp)^2+b(x+kp)+c\\ &=&(ax^2+bx+c)+(2ax+b)kp+ak^2p^2\\ &=&((2ax+b)k+m)p+ak^2p^2 \end{eqnarray*} We see that $y$ satisfies $(2)$ if and only if $$(2ax+b)k+m\equiv0\pmod{p}.$$ Now because $2ax+b\not\equiv0\pmod{p}$ we are be done by taking $$k\equiv-m(2ax+b)^{-1}\pmod{p}.$$