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

$2x^2 - 7 \cong 2x^2 + 4 - 11 \cong 2(x^2+2)\pmod {11} \to x^2+2 = 0\pmod {11} \to x^2=-2\pmod {11}=9\pmod {11}\to x^2-9 = 0\pmod {11}\to (x-3)(x+3)=0\pmod {11}\to x-3=0\pmod {11} \text{  or } x+3 = 0\pmod {11}\to x=3,8 $ $\text{since}$ $0\leq x < 11$.
