Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$

My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv1$, $x\equiv1\pmod7$ is a solution.
We proceed to lift the solution $\mod 7^2$
$f'(x) = 11x^{10} + 8x^7 +5$, we have $f(1) = 7$ , $f'(x) = 24$
Since $7$ can not divide $f'(1)$, we need to solve $24t \equiv 0 \mod 7$
we get $t \equiv 0 \pmod 5$
and then ..................... ?????

Why not use Hensel's Lemma ?
First, let us find a solution modulo $\,7\,$ of $\,f(x)=x^{11}+x^8+5\pmod 7\,$.
Clearly, $\,r=1\,$ is a root. Now, define
$$t:=-\frac{f(1)}{7}f'(1)^{-1}=-(11+8)^{-1}=(-19)^{-1}=30^{-1}=18\pmod{49}\Longrightarrow $$
$$s:=1+18\cdot 7=127=29\pmod{49}\,\,\,\text{is a root of}\,\,\,f(x)\pmod{49}$$
What is cool in this method is that it allows you find roots modulo $\,7^2\,,\,7^3\,\ldots\,,\,7^k\,$ , for any $\,k\in\Bbb N\,$ , as long as you have one simple root modulo $\,7\,$ (why does it have to be "simple"?)
