Solve $3x^2+6x+5\equiv 0\pmod{89^2}$.
To do this, I first solved $3x^2+6x+5\equiv 0\pmod{89}$.
This has a solution since $3x^2+6x+5 = 3(x+1)^2 + 2$ and $3(x+1)^2 \equiv -2 \pmod{89}$ has a solution since $(3,89)=1$, we can check whether $(x+1)^2 = 3^{-1}(-2)\pmod{89}$ which is equivalent to calculating $\left(\dfrac{3^{-1}(-2)}{89}\right)$. The value is 1, so this is a quadratic residue. Since we have a number that when squared gives us this, we have a solution to the polynomial congruence mod 89. Since we have 1, we have 2 since solutions come in pairs $\{\pm x_0\}\pmod{89}$.
Here is my question: without finding the residue, how can I use Hensel's lemma to see whether I can lift the solutions to solutions modulo $89^2$?
I hope someone can help. Thank you.