Find a(n) with characteristic equation

Question

Find a(n) with characteristic equation of recurrence: a(1)=3, a(2)=10, a(n)=6a(n-1)-3a(n-2) .

Started from equation t^2-6t+3=0 then find t1,t2 then take them in: a(n)=Ct1+ Dt2 then C and D have to find from a(1)=3 and a(2)=10 but I'm stuck here.

Answer 1

$t^2-6t+3=0 \implies t=3 \pm \sqrt{6}$. Thus $$a_n=C(3+\sqrt{6})^n + D(3-\sqrt{6})^n.$$ Now substitute $n=1$, to get $$a_1=3=C(3+\sqrt{6})^{\color{red}{1}}+D(3-\sqrt{6})^{\color{red}{1}}.$$ Likewise with $n=2$ we get $$a_2=10=C(3+\sqrt{6})^{\color{red}{2}}+D(3-\sqrt{6})^{\color{red}{2}}=C(15+6\sqrt{6})+D(15-6\sqrt{6}).$$ So now solve for $C$ and $D$ from the following set \begin{align*} C(3+6\sqrt{6})+D(3-6\sqrt{6})&=3\\ C(15+6\sqrt{6})+D(15-6\sqrt{6})&=10. \end{align*} I believe \begin{align*} C&=\frac{1}{12}(16-5\sqrt{6})\\ D&=\frac{1}{12}(16+5\sqrt{6}) \end{align*}