How to solve this equation with matrices can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.

Let $$X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies X^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{bmatrix}=\begin{bmatrix} 6 & 2\\ 3 & 7 \end{bmatrix}.$$
$$\implies a^2+bc=6~~~(1), b(a+d)=2~~~(2), c(a+d)=3~~~(3), d^2+bc=7~~~(4)$$
From (2) and (3) $b/c=2/3$, let $b=2k, c=3k$. From (1) and (4), we get $a^2-d^2=-1 \implies (a-d)(a+d)=-1$. Again from (2) and (3) we get $a+d=1/k, a-d=-1/k \implies a=(1/k-k)/2, d=(1/k+k)/2$. Finally,inserting $a,b,c$ in (1), we get
$$25k^4-26k^2+1=0 \implies k=\pm 1,\frac{\pm 1
}{5}$$
For $k\pm 1$, we get
$$X=\pm \begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix}~~~(1)$$
For $k=\pm\frac{1}{5}$, we get
$$X=\pm \frac{1}{5} \begin{bmatrix} 12 & 2 \\ 3 & 13 \end{bmatrix}~~~(2)$$
This sheer brute force method gives 4 matrices. one can check their correctness by squaring them.
