The given problem is a matrix equation that needs to be solved. The equation is written as:

$X \cdot \begin{pmatrix} 3 & -4 \\ -5 & 6 \end{pmatrix} = \begin{pmatrix} 5 & 1 \\ 2 & 0 \end{pmatrix}$

To solve for $X$, we need to multiply both sides of the equation by the inverse of the matrix on the left. The steps are as follows:

1. Find the inverse of matrix $A = \begin{pmatrix} 3 & -4 \\ -5 & 6 \end{pmatrix}$.

The inverse of a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by:

$A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

In this case, $a = 3$, $b = -4$, $c = -5$, and $d = 6$.

The determinant is:

$\text{det}(A) = (3)(6) - (-4)(-5) = 18 - 20 = -2$

Thus, the inverse of $A$ is:

$A^{-1} = \frac{1}{-2} \begin{pmatrix} 6 & 4 \\ 5 & 3 \end{pmatrix} = \begin{pmatrix} -3 & -2 \\ -\frac{5}{2} & -\frac{3}{2} \end{pmatrix}$

2. Multiply both sides of the equation by $A^{-1}$ on the left:

$X = \begin{pmatrix} 5 & 1 \\ 2 & 0 \end{pmatrix} \cdot A^{-1}$

Now, perform the matrix multiplication:

$X = \begin{pmatrix} 5 & 1 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} -3 & -2 \\ -\frac{5}{2} & -\frac{3}{2} \end{pmatrix}$

3. Multiply the matrices:

$X = \begin{pmatrix} (5)(-3) + (1)\left(-\frac{5}{2}\right) & (5)(-2) + (1)\left(-\frac{3}{2}\right) \\ (2)(-3) + (0)\left(-\frac{5}{2}\right) & (2)(-2) + (0)\left(-\frac{3}{2}\right) \end{pmatrix}$

Simplifying: