A simple residue

By the Law of Quadratic Reciprocity, $x^2 = 11 (mod 43)$ is solvable iff $x^2 = 43 (mod 11)$ is not solvable.

43 = -1 (mod 11), so this reduces to $x^2 = -1 (mod 11)$

A simple check reveals that mod 11, squares have residues of 0, 1, 4, 9, and 5.

(If you're still not sure, $21^2 = 441 = 11(mod 43)$ , so...)

$\left( \dfrac{11}{43}\right)L = \left(\dfrac{-43}{11}\right)L = \left( \dfrac{1}{11}\right)L = 1^5 = 1$

Since the value of the Legende Symbol is 1, 11 is a quadratic residue modulo 43, and therefore it is solvable.