If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x  \equiv -3 $ (mod 11)
$-1(x^2 + x)  \equiv -3 $ (mod 11)
$(x^2 + x)  \equiv 3 $ (mod 11)
-Not sure if I am heading the right way, once again any help is appreciated.

In $\mathbb F_{11}$, one has $$7x^2 - 4x + 1= 0\iff-4x^2-4x+1=0\iff x(x+1)=\frac {1}{2^2}=2^8=3$$
For all $x\in\mathbb F_{11}$ we have $x(x+1)\ne 3$ (successively one has $0,2,6,1,9,8,9,1,6,2,0$) 
Hence there is not solution.
