Alphabet with 6 vowels and 12 consonants, find the amount of words without two consonants in a row.

$$a_n=6a_{n-1}+6\cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial $$ r^2-6r-6\cdot 12 $$ which factors as $$ (r-12)(r+6) $$ Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore $$ a_n=A\cdot 12^n+B(-6)^n $$ and you can do the rest.