Since the question has already been settled nicely, let me demonstrate an unconventional way to find the tangent of a circle through a given point on the circle.
There are three things that allow us to do what we will be doing in the sequel:
Any point can be thought of as a "point-circle"; that is, a circle of radius $\textbf 0$.
One can always produce the equation for the radical line of two circles.
The radical line of two tangent circles is their common tangent line.
Here, now, is the procedure. We are given the point $(-2,5)$ on the circle $x^2+y^2+6x-4y+3=0$. Constructing the equation of a point-circle is a snap (note that as always, we negate the coordinates of the center when inserting them into the equation of a circle):
which in expanded form, looks like
From the equivalence of the tangent line and the radical line in this case that I mentioned earlier, all we have to do to find the tangent line to your circle is to find the radical line of the given circle and the point-circle. (Certainly, a point circle lying on another circle is necessarily tangent to it.) The equation of the radical line is easily produced by just subtracting the two given equations; thus,
or, after simplification and solving for $y$,
is the sought radical line, which is also the tangent line of the circle $x^2+y^2+6x-4y+3=0$ through $(-2,5)$.