Congruency by completing the square $x^2+x+1\equiv 0\mod 49$
We have the ring isomorphism $\mathbb{Z}/49\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$.
Consider $x^2+x+1\equiv 0\mod 7$
I usually solve these polynomial congruencies using the 'complete the square' method. How would I go about doing that in this case?
EDIT:
$x^2+x+1\equiv x^2-6x+8\equiv(x-3)^2\equiv 1\mod 7$
I get solutions $x=4$ and $x=2$ $\mod 7$.
Now I tried to use Hemel's lifting, so,
We have $f(x)=x^2+x+1$ thus $f'(x)=2x+1$.
In the case of $x\equiv 2\mod 7$, we have $f(2)=7$, $f'(2)=5$; as $7\nmid f'(2)$, we can lift to a unique solutions by solving $f'(2)\equiv -\frac{f(2)}{7}\mod 7$, i.e. $5t\equiv -1\mod 7$.
Clearly $t\equiv 4\mod 7$, leading to the solution $x\equiv 2+4\cdot 7\equiv 30\mod 49$.
In the case of $x\equiv 4\mod 7$, we have $f(4)=21$, $f'(4)=9$; as $7\nmid f'(4)$, we can lift to a unique solutions by solving $f'(4)t\equiv -\frac{f(4)}{7}\mod 7$ or $9t\equiv -\frac{9}{7}\mod 7$...and this is where I'm stuck, assuming I'm even doing it correctly.

A start: Equivalently, we solve $4x^2+4x+4\equiv 0\pmod{49}$, that is, $(2x+1)^2+3\equiv 0\pmod{49}$.
