Hi, so I am trying to find one solution of the polynomial congruence:

3x2+2x+100 0(mod343)

So I know that this is equivalent to 3x2+2x+100 0(mod73), so I did 3x2+2x+100 0(mod7), and plugged in the values 1-6 to find that we get solutions x = 1,3.

From here on I am a little confused. I know it is something to do with finding the derivative f'(x) and putting the values in and using Hensel's Lemma but im really not sure what the method is!

Thanks!