1. The roots $\alpha, \beta$ satisfy $5x^2+6x-2=0.$ So each root $r$ satisfies $5r^2+6r-2=0.$

2. Rearranging, $5r^2+6r=2.$ Multiply both sides by $r^{n-2}$ (for $n\ge 2$): $5r^n+6r^{n-1}=2r^{n-2}.$

3. This is true for both roots $\alpha$ and $\beta$. Adding for $r=\alpha$ and $r=\beta$, $5(\alpha^n+\beta^n)+6(\alpha^{n-1}+\beta^{n-1})=2(\alpha^{n-2}+\beta^{n-2}).$ Using $S_n=\alpha^n+\beta^n$, $5S_n+6S_{n-1}=2S_{n-2}.$

4. Put $n=6$: $5S_6+6S_5=2S_4.$ This matches option A.

5. Checking the other options:

• B says $5S_6+6S_5+2S_4=0$, but from the recurrence, $5S_6+6S_5=2S_4,$ so this would give $4S_4=0$, not generally true.
• C and D do not match the derived recurrence.

Therefore the correct option is: $\boxed{A}$

Comparison with stored answer: Stored correct answer is A, which agrees with the derived result.