Linearizing systems about critical points. $$\def\q{\begin{pmatrix}}\def\p{\end{pmatrix}}\def\l{\lambda}\def\f{\frac{\sqrt{11}}{2}}$$

Find all the critical points of the following systems and derive the linearised system about each
  critical point. Use this to determine the nature of the critical points and whether they are stable
  or unstable.
What about say:
  $$\q y_1' \\ y_2'\p = \q 2y_1 + y_2\\ -5y_1 + 5 \p$$
Here I have a form I haven't before worked with, being that I can't just look at the eigenvalues to find type of critical point, since I can't put it in the form $\vec{y'} = A\vec{y}$ where $A$ is some coefficient matrix. What do I do here?

Form I have worked with
$$\vec{y'} = \q 2 & 1 \\ -5 & 5 \p \q y_1 \\ y_2 \p$$
$$\q 2-\l &1 \\-5&5-\l \p = 0$$
$$=(2-\l)(5-\l)+5 = 0$$
$$=\l -7\l + 15 = 0$$
$$=\frac{7 \pm \sqrt{(-7)^2 - 60}}{2}$$
$$=\frac{7\pm \sqrt{11}i}{2}$$
$$\l =3.5 \pm \frac{\sqrt{11}}{2}$$
$$\l=3.5 + \f, \l = 3.5 - \frac{\sqrt{11}}{2}$$
Since we have complex conjugate eigenvalues with non-zero real part, we know this is a spiral critical point at the origin. I could then sub in a few vectors to determine if it was asymptotically un/stable.

The system is $y'=Ay+u$, where $A= \begin{bmatrix} 2 & 1 \\ -5 & 0 \end{bmatrix}$, and $u = \begin{bmatrix} 0 \\ 5 \end{bmatrix}$.
The critical points are where $y'=0$, this gives $y^* = -A^{-1} u = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$.
The linearised system can be written as $(\upsilon+y^*)' = \upsilon' = A(\upsilon+y^*) +u = A \upsilon$.
Since the eigenvalues of $A$ are $1 \pm 2 i$, it is clear that the linearised system is unstable.
