Three cases of order-two recurrences

1. Two real roots. $r^2-5r+6=0$ gives $r=2,3$. So $u_n=A2^n+B3^n$ with $A+B=1$ and $2A+3B=0$, hence $B=-2$, $A=3$:
$$u_n=3\cdot2^n-2\cdot3^n.$$

2. Double root. $r^2-4r+4=(r-2)^2$, so $u_n=(A+Bn)2^n$. $u_0=1$ gives $A=1$ and $u_1=2(1+B)=0$ gives $B=-1$:
$$u_n=(1-n)2^n.$$

3. Complex roots. $r^2-2r+2=0$ has roots $1\pm i=\sqrt2\,e^{\pm i\pi/4}$. So $u_n=2^{n/2}\left(A\cos\frac{n\pi}4+B\sin\frac{n\pi}4\right)$. $u_0=0$ gives $A=0$, and $u_1=\sqrt2B\sin\frac\pi4=B=1$:
$$u_n=2^{n/2}\sin\frac{n\pi}4.$$

Checks. (1) $u_2=-6=12-18$. (2) $u_2=-4=(1-2)\cdot4$. (3) $u_2=2=2\sin\frac\pi2$ and $u_3=4-2=2=2^{3/2}\sin\frac{3\pi}4$.

Remark. In case 3 the sequence rotates and grows at the same time: it oscillates with an amplitude $2^{n/2}$.