Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.

I feel that this may be too elementary, is this the correct way to solve this?

Indeed, it is wrong. $49 \mid (x+2)^2+14$ implies $7\mid (x+2)^2$, and in particular $7\mid x+2$. But this means that $49\mid (x+2)^2$, and by difference $49$ should divide $14$ too, which is false.
