The relation to solve is this:
$$ a_{n} = 7a_{n-1} - 10a_{n-2}, a_{0} = 5, a_{1} = 16$$
So
$$ a_{2} = 62, a_{3} = 274, ...$$
So I thought I was supposed to be able to do this to solve:
$$ x^n = 7x^{n-1} - 10x^{n-2}; x^2 = 7x - 10, x^2 - 7x + 10 = 0; x = 2,5 $$
But I can't plug those in? a_2 would be 7(5)^1 - 10(2)^0 != 62, nor the other way around (switching the positions of the injected roots)... a_3 same sad story. Am I close or not at all?
So what we have is:
$a_{0} = A(5)^{0} + B(2)^{0} = A + B = 5$
$a_{1} = 5A + 2B = 16$
So $3A = 6 \implies A = 2$. Similarly, $B = 3$.
So your solution is: $a_{n} = 2 * 5^{n} + 3 * 2^{n}$.