$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:

$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.

I don't really know what to do from here. If I substitute 
$$R_n = 3(2^n)-4(5^n)$$
into 
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up getting 
$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$
Dont know what to do...
EDIT: Thanks to Zev, what I did was:
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$

Having rewritten the recurrence, we get an equation:
$$10R_{n-2}-7R_{n-1}+R_n=0$$
that can easily be solved.
The roots of a characteristical polymomial $10-7k+k^2=0$ are $k_1=2,k_2=5$.
So general solution is $R_n=c_12^{n}+c_25^{n}$, where $c_1$ and $c_2$ are arbitrary constants.
