Please note this is my first time answering, so help me improve by keeping comments constructive
This is the characteristic polynomial method for finding a closed form expression of a recurrence relation, similar and dovetailing other answers:
given: $f(0)=v_1$, $f(1)=v_2$, ..., $f(d-1)=v_{d-1}$ and $a_df(n) + a_d−1f(n − 1) + · · · + a_0f(n-d) = 0$ for all $n ≥ 0$
Note, you likely need to rewrite the recurrence relation, for example, as your recurrence relation presents $x_n = 11x_{n-1} - 30x_{n-2}$ becomes $x_n-11x_{n-1}+30x_{n-2}=0$
The characteristic polynomial of this recurrence relation is of the form:
$q(x) = a_dx^d + a_{d−1}x^{d−1} + · · · + a_1x + a_0$
Now it's easy to write a characteristic polynomial using the coefficents $a_d$,$a_{d-1}$, ..., $a_0$:
$q(r)=r^2-11r+30$
Since $q(r) = 0$, the geometric progression $f(n) = r^n$ satisfies the implicit recurrence.
IF the roots of the characteristic equation are distinct, $f(n) = λ_1r_1^n + λ_2r_2^2 + · · · + λ_dr_d^n$, where $λ_1, . . . , λ_d$ are arbitrary complex numbers.
In this case, we have:
$q(r)=r^2-11r+30$
$q(r)=(r-5)(r-6)$
$r_1=5$ and $r_2=6$
So the general solution is:
$x_n=λ_15^n + λ_26^n$
Use the initial values to get two equations with two unknowns:
$f(0)=λ_15^0 + λ_26^0$
$4=λ_1 + λ_2$
$f(1)=λ_15^1 + λ_26^1$
$23=5λ_1 + 6λ_2$
Solving this, we have $λ_1 = 1$ and $λ_2=3$, so the solution is
$x_n=5^n + 3*6^n$
IF the roots are NOT distinct, we need to think a bit further, but we'll leave that for another time.