Solving a linear recurrence with a multiplicity of two

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I was given this problem and I am trying to figure out where I go wrong

solve the linear recurrence:

$f(0) = 0$, $f(1) = 0$, $f(2) = 18$,

$f(n) = 3f(n − 1) − 4f(n − 3)$

Here is what I have so far: Assume: $f(n)=x^n$

$x^n=3x^{n-1}-4x^{n-3}$

$x^3-3x^{2}-4=0$

This factors to:

$(x-2)^2(x+1)$

$x=2$with a multiplicity of $2$ $x=-1$

I think this is where I am having some problems:

$f(n)=ax^n+bnx^n+cx^n$

$f(n)=a2^n+bn2^n+c(-1)^n$

Now I have to solve for $a b c$

EDIT

$f(0) = 0$:$f(0)=a2^0+b(0)2^n+c(-1)^0=0$ Simplified: $a-c=0$

$f(1) = 0$: $f(1)=a2^1+b(1)2^n+c(-1)^1=0$ Simplified: $2a+4b-c=0$

$f(2) = 18$: $f(2)=a2^2+b(2)2^n+c(-1)^2=18$ Simplified: $4a+8b+c=18$

$a=6$, $b=-3/2$, $c=6$

can someone tell me if I made a mistake so far because I keep getting the wrong answer