$$a_n=4a_{n-1}+5a_{n-2},\quad a_1=2,a_2=6$$
$$x^2-4x-5=0$$
$$x=-2+i,-2-i$$(complex roots)
as per the quadratic equation for the roots, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Then what is the formula ?
thank for confirming the roots just need one more clarification, after substituting these roots my a_n=C_1(-1)^n+C_2(5)^n =>a_1=C_1(-1)^1+C_2(5)^1 =>a_1=-C_1+5C_2 similarly for a_2=C_1(-1)^2+C_2(5)^2 =>a_2=C_1+25C_2 If this solution correct.
Beside the fact that the roots you gave in your post are wrong, the solution to your problem is fully described in the link lab bhattacharjee provided in his comment.
So, learn (if still required) the method and apply it. You should arrive to $$a(n)=\frac{2}{15} \left(2\ 5^n-5 (-1)^n\right)$$