Mary: You may consider your equation for $n+2$ instead of $n+1$; this gives us $$ a(n+2)=2a(n+1)-a(n)-1.$$ Subtract the first equation from this one. We obtain $$\begin{array}{rl}a(n+2)-a(n+1)&=(2a(n+1)-a(n)-1)-(2a(n)-a(n-1)-1)\\&=2a(n+1)-3a(n)+a(n-1),\end{array}$$ or $$a(n+2)=3a(n+1)-3a(n)+a(n-1).$$
In general, if your equation is "almost homogeneous" but for a constant, the same trick of subtracting consecutive instances of the recurrence gives us a homogeneous equation.
The associated characteristic equation is now $x^3-3x^2+3x-1=0$, or $(x-1)^3=0$. This means that $a(n)=A+Bn+Cn^2$ for some constants $A,B,C$.
In general, if the equation has the form $(x-r)^k(x-b)^s\dots=0$, then the general solution will have the form $$a(n)=(A+Bn+\dots+C n^{k-1})r^n+(D+En+\dots+F n^{s-1})b^n+\dots$$ (in the most studied case, the characteristic equation has no repeated roots, but as in your example here, repeated roots may occur, so we need this more general version).
To find $A,B,C$ in our equation for $a(n)=A+Bn+Cn^2$, we use the given initial conditions. Are you sure the condition $a(10)=0$ is correct, rather than $a(1)=0$?
With $a(1)=0$, we have $0=A+B+C$. Also, $a(0)=0$, so $A=0$. Finally, $a(2)=2a(1)-a(0)-1=-1$, using the original recurrence, so $A+2B+4C=-1$. This easily gives us $A=0$, $B=1/2$, and $C=-1/2$, or $$ a(n)=\frac{n-n^2}2. $$
Note that we needed to compute $a(2)$ before we could find $A,B,C$. This is because the original equation was not homogeneous, so even though it is of second order, we needed an additional initial condition to the two given to us. (Note the homogeneous equation we obtained is of third order, needing 3 initial conditions.)
Unfortunately, I do not know of a decent reference on recurrence relations at an elementary level (I know of one, in Spanish, a translation of a small Russian booklet published by Mir eds. years ago). There are several standard approaches which might not be as elementary as you may want, but you may enjoy looking at them anyway: Using linear algebra or generating functions. For the latter, a good reference is "generatingfunctionology",
http://www.math.upenn.edu/~wilf/DownldGF.html
and for the former a good reference is Volume II of Apostol's Calculus book.
Hmm... With $a(10)=0$, we can proceed as follows: We have $$a(n)=A+Bn+Cn^2$$ and so $$A=0$$ (since $a(0)=0$) and $$10B+100C=0$$ (since $a(10)=0$ and $A=0$). We also have $$a(n+2)=2a(n+1)-a(n)-1,$$ or
$$ B(n+2)+C(n+2)^2=2B(n+1)+2C(n+1)^2-Bn-Cn^2-1, $$ which simplifies to $2C=-1$, or $$C=-1/2.$$ Then $$B=5,$$ from the equation for $a(10)=0$, and $$ a(n)=5n-\frac{n^2}2. $$
The characteristic equation is the one that a number $\lambda$ should satisfy in order for the geometric series $(\lambda^n)_{n\in\mathbf N}$ to be a solution of the recurrence relation. Another interpretation is that if you interpret the indeterminate $s$ as a left-shift of the sequence (dropping the initial term and renumbering the renaming terms one index lower), then the characteristic equation gives the lowest degree monic polynomial that when applied to this shift operation kills all sequences satisfying the recurrence. In the case of Fibonacci recurrence, applying $s^2-s-1$ to a sequence $A=(a_i)_{i\in\mathbf N}$ gives the sequence $(a_{i+2}-a_{i+1}-a_i)_{i\in\mathbf N}$, which is by definition identically zero if (and only if) $A$ satisfies the recurrence.
A different but related polynomial that is of interest is obtained by reversing the order of the monomials (giving a polynomial starting with constant term $1$), so for Fibonacci it would be $1-X-X^2$. This polynomial $P$ has the property that the formal power series $F=\sum_{i\in\mathbf N}a_iX^i$ associated to a sequence satisfying the recurrence, when multiplied by $P$ gives a polynomial $R$ (with $\deg R<\deg P$), in other words all terms from the index $\deg P$ on are killed. This is basically the same observation as for the shift operation, but the polynomial $R$ permits describing the power series of the sequence, including its initial values, formally as $F=\frac RP$. For the Fibonacci sequence one finds $R=X$ so its formal power series is $F=\frac X{1-X-X^2}$.
Best Answer
Associated with the recurrence $f(n)=-3f(n-1)+4f(n-2)$ is a so-called characteristic equation, $x^2=-3x+4$. Its coefficients are the same as the coefficients of the recurrence, and the powers of $x$ are chosen so that the smallest exponent is $0$, associated with the smallest argument of $f$, which in this case is $n-2$; the exponents then increase in step with the arguments of $f$, so that exponent $1$ goes with $(n-2)+1=n-1$, and exponent $2$ goes with $(n-2)+2=n$.
Now solve the auxiliary equation: $x^2+3x-4=0$, $(x+4)(x-1)=0$, $x=-4$ or $x=1$.
There is a general theorem that says that when the roots are distinct, as they are here, the general solution to the recurrence has the form
$$f(n)=Ar_1^n+Br_2^n\;,$$ where $r_1$ and $r_2$ are the two roots. Thus, for this recurrence the general solution is $$f(n)=A(-4)^n+B\cdot1^n=A(-4)^n+B\;.\tag{1}$$
$(1)$ gives all solutions to the recurrence $f(n)=-3f(n-1)+4f(n-2)$, for all possible initial values of $f(1)$ and $f(2)$. To determine which values of $A$ and $B$ correspond to your particular initial values, substitute $n=1$ and $n=2$ into $(1)$. For $n=1$ you get $$1=f(1)=A(-4)+B\;,$$ and for $n=2$ you get $$2=f(2)=A(-4)^2+B\;.$$
Now you have a system of two equations in two unknowns,
$$\left\{\begin{align*} &-4A+B=1\\ &16A+B=2\;. \end{align*}\right.$$
Solve this system for $A$ and $B$, substitute these values into $(1)$, and you have your general solution. (I get $A=\frac1{20}$ and $B=\frac65$.)
Note that if the the roots $r_1$ and $r_2$ of the characteristic equation are equal, say $r_1=r_2=r$, the general solution is a little different:
$$f(n)=Ar^n+Bnr^n\;.$$ However, you solve for the particular $A$ and $B$ in the same way.