I think you just have to show that $b_n = 5\cdot 3^n + 7\cdot 2^n$ fulfills the recurrence relation and initial conditions for every instance $n$.
As the relation involves up to two prior sequence elements I would use strong induction.
You start with direct proofs for $b_0$, then $b_1$ and the induction from $n=2$.
Update: It seems I need to provide a full solution:
The initial conditions hold by evaluation and comparison.
$$
S(0): b_0 = 12 \wedge
b_0 = 5\cdot 3^0 + 7\cdot 2^0 = 5 + 7 = 12 \\
S(1): b_1 = 29 \wedge
b_1 = 5\cdot 3^1 + 7\cdot 2^1 = 15 + 14 = 29 \\
$$
Where the $\wedge$ means logical "and".
The statement for $n$, $n\ge 2$ is
$$
S(n): b_n = 5 b_{n-1} - 6 b_{n-2} \wedge
b_n = 5\cdot 3^n + 7\cdot 2^n
$$
Base case $n=2$:
$$
S(2):b_2 = 5 b_1 - 6 b_0 \wedge b_2 = 5 \cdot 3^2 + 7 \cdot 2^2
$$
This evaluates to
$$
S(2): b_2 = 5 \cdot 29 - 6 \cdot 12 = 145-72=73 \wedge b_2 = 45 + 28 = 73
$$
which is a true statement.
Induction Step:
Assuming $S(k)$ is true for $k\in\{2,\dotsc, n\}, n\ge 2$ we have
$$
S(k+1): b_{k+1} = 5 b_k - 6 b_{k-1} \wedge
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
where
\begin{align}
b_{k+1}
&= 5 b_k - 6 b_{k-1} \\
&= 5 \left( 5\cdot 3^k + 7\cdot 2^k \right) -
6 \left( 5\cdot 3^{k-1} + 7\cdot 2^{k-1} \right)
\end{align}
The first term in parentheses is justified by the truth of $S(k)$.
The second term needs a case distinction:
For $k=2$ it relies on the truth of the fulfillment of the base condition $b_1$, otherwise we can build on the true statements $S(k-1)$.
Then:
$$
\begin{align}
b_{k+1}
&= (25 - 10) 3^k + (35-21) 2^k \\
&= 15\cdot 3^k + 14\cdot 2^k \\
&= 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
\end{align}
$$
so we end up with
$$
S(k+1): b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1} \wedge
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
which is a true statement.
By the principle of strong induction we conclude the truth of $S(n)$ for all integer $n$ with $n>=2$.
Adding the truth of the initial conditions we can claim the truth of
$S(n)$ for all integer $n$ with $n \ge 0$.
Now
\begin{align}
S(0)&: b_0 = 12 \wedge
b_0 = 5\cdot 3^0 + 7\cdot 2^0 \\
S(1)&: b_1 = 29 \wedge
b_1 = 5\cdot 3^1 + 7\cdot 2^1 \\
S(n)&: b_n = 5 b_{n-1} - 6 b_{n-2} \wedge
b_n = 5\cdot 3^n + 7\cdot 2^n \quad (n \ge 2)
\end{align}
means that for all integer $n$ with $n\ge 0$ the sequence $b_n = 5\cdot 3^n + 7\cdot 2^n$ is the solution of the recurrence relation.
