Show that n²+2n+12 is not a multiple of 121 for any integer n.

Dear Student,
We first make the assumption that n²+2n+12 is a multiple of 121.
i.e n²+2n+12=121*k(Where k is any positive integer)
i.e n²+2n+12-121*k=0.
Formula Method for Quadratic Equation$a{x}^2+bx+c=0$ is x=$-b\pm \sqrt{b^2-4ac}/2a$
Now, solving for n using the Formula Method, 
n= $-2\pm \sqrt{484*k-44}/2$
Given that n$n$ is an integer, so (484⋅k)−11−−−−−−−−−−−√$\sqrt{(484\cdot k)-44}$ should be an integer
But it can also be written as:-
i.e n=$-2\pm \sqrt{44}\left(\sqrt{11k-1}\right)/2$
But this value cannot be an integer as the vaue of $\sqrt{44}$is irrational.
Therefore our assumption that 121 is a multiple of the equation is wrong.
Hence the following statement is proved.
Regards.