How to turn the recurrence $a_n=2a_{n−1}−a_{n−2}$, with $a_0=−1$ and $a_1=1$ into a quadratic, to find the closed form? I am having trouble understanding how to turn the recurrence into a quadratic equation. What principles are involved, rules etc.?

You can write the recurrence with matices:
$$
\begin{pmatrix}
a_{n-1} \\
a_{n}   \\
\end{pmatrix}
=
\begin{pmatrix}
 0 & 1 \\
-1 & 2 \\
\end{pmatrix}
\begin{pmatrix}
a_{n-2} \\
a_{n-1} \\
\end{pmatrix}
$$
Then,
$$
\begin{pmatrix}
a_{n-1} \\
a_{n}   \\
\end{pmatrix}
=
\begin{pmatrix}
 0 & 1 \\
-1 & 2 \\
\end{pmatrix}^{n-1}
\begin{pmatrix}
-1 \\
1 \\
\end{pmatrix}
$$
You can Jordan-decompose the matrix in the power as $M = S J S^{-1}$
$$
\begin{pmatrix}
 0 & 1 \\
-1 & 2 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 \\
1 &  0 \\
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
 0 & 1 \\
-1 & 1 \\
\end{pmatrix}
$$
To do that manually, you can use the characteristic polynomial, which probably is what you meant with quadratic equation.
$$
\det(M - \lambda \mathbb{1})
=
\begin{vmatrix}
-\lambda & 1 \\
-1 & 2-\lambda \\
\end{vmatrix}
= \lambda^2 -2 \lambda + 1
= (\lambda^2 - 1)^2
= 0
$$
That means that there is only the eigenvalue $\lambda = 1$. Then you find eigenvectors and etcetera. Probably there is some answer which already explains how to do the decomposition in detail.
Once you have the decomposition, it's easy to see that
$$
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}^{n-1}
=
\begin{pmatrix}
1 & n-1 \\
0 & 1 \\
\end{pmatrix}
$$
Therefore,
$$
\begin{pmatrix}
 0 & 1 \\
-1 & 2 \\
\end{pmatrix}^{n-1}
=
\begin{pmatrix}
1 & -1 \\
1 &  0 \\
\end{pmatrix}
\begin{pmatrix}
1 & n-1 \\
0 & 1   \\
\end{pmatrix}
\begin{pmatrix}
 0 & 1 \\
-1 & 1 \\
\end{pmatrix}
$$
And finally,
$$
\begin{pmatrix}
a_{n-1} \\
a_{n}   \\
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 \\
1 &  0 \\
\end{pmatrix}
\begin{pmatrix}
1 & n-1 \\
0 & 1   \\
\end{pmatrix}
\begin{pmatrix}
 0 & 1 \\
-1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-1 \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
2n-3 \\
2n-1 \\
\end{pmatrix}
$$
The general term is
$$a_n = 2n-1$$
