Calculate the determinant of the left-hand side matrix.

The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$.
For the left-hand side:

$$\begin{vmatrix} -2n+3 & n+1 \\ 2n & n \end{vmatrix} = (-2n+3)(n) - (n+1)(2n)$$

Expand the terms:

$$(-2n^2 + 3n) - (2n^2 + 2n)$$ $$-2n^2 + 3n - 2n^2 - 2n$$ $$-4n^2 + n$$

For the right-hand side:

$$\begin{vmatrix} -2n+5 & 4n \\ -n+2 & 3n \end{vmatrix} = (-2n+5)(3n) - (4n)(-n+2)$$

Expand the terms:

$$(-6n^2 + 15n) - (-4n^2 + 8n)$$ $$-6n^2 + 15n + 4n^2 - 8n$$ $$-2n^2 + 7n$$

$$-4n^2 + n = -2n^2 + 7n$$

Move all terms to one side to form a quadratic equation:

$$0 = -2n^2 + 7n + 4n^2 - n$$ $$0 = 2n^2 + 6n$$

Factor out $2n$:

$$2n(n + 3) = 0$$

This gives two possible solutions for $n$:

$$2n = 0 \implies n = 0$$ $$n + 3 = 0 \implies n = -3$$

The solutions for $n$ are 0 and $-3$.

$$n = 0, -3$$