Solve $x^2+5x+24=0$ mod $36$ 
Solve $x^2+5x+24=0$ mod $36$

So I have $x^2+5x+24\equiv 0$ mod $2^2\cdot 3^2$
SO I believe I need $x^2+x\equiv 0$ mod $2$ and $x^2+2x\equiv 0$ mod $3$
Which I determined that $0,1$ both work.
Then $f^\prime(x)=2x+5$
$f^\prime(0)=5$ and $f^\prime(1)=7$. So by hensel lemma there are solutions mod $4$
I'm not sure what to do now, am I supposed to lift to solutions mod $2^2$ and mod $3^2$?
I have $f(0+2t)=f(0)+2tf^\prime(0)\equiv 0 $ mod $4$
is $f(2t)=0+2tf^\prime(0)=2t(5)\equiv 0 $ mod $4$
then $10t\equiv 0$ mod $4$
which says $t=0$ which seems fine because $f(0)\equiv 0$ mod $4$ but I'm not sure I'm doing this right.
Im not sure how to deal with hensel lifting on numbers which aren't prime powers.

You could solve this without Hensel lifting.
$x^2+5x+24\equiv0\bmod36\implies x^2+x\equiv0\bmod4$ and $x^2+5x+6\equiv0\bmod9$.
This means $x(x+1)\equiv0\bmod4$ and $(x+2)(x+3)\equiv0\bmod9$.
Note that $4|x(x+1)\implies 4|x$ or $4|x+1$, and $9|(x+2)(x+3)\implies 9|x+2$ or $9|x+3$,
because it is not possible for $2|x$ and $2|x+1$ or $3|x+2$ and $3|x+3$.
Anyway, can you take it from here?
