Solving $2x^2 \equiv 7 \pmod{11}$ I'm doing some previous exams sets whilst preparing for an exam in Algebra.
I'm stuck with doing the below question in a trial-and-error manner:
Find  all $ x \in \mathbb{Z}$ where $ 0 \le x \lt 11$ that satisfy $2x^2 \equiv 7 \pmod{11}$
Since 11 is prime (and therefore not composite), the Chinese Remainder Theorem is of no use? I also thought about quadratic residues, but they don't seem to solve the question in this case.
Thanks in advance

Hints: 


*

*$6\cdot2=1\pmod{11}$

*$6\cdot7=9\pmod{11}$

*$ab=0\pmod{11}\iff (a=0\pmod{11}\vee b=0\pmod{11})$

