Find the asymptotes to the hyperbola $3x^2+2xy-y^2+8x+10y+14=0$ 
The hyperbola is given with the following equation: $$3x^2+2xy-y^2+8x+10y+14=0$$
  Find the asymptotes of this hyperbola. ($\textit{Answer: }$ $6x-2y+5=0$ and $2x+2y-1=0$)

In my book, it is said that if the hyperbola is given with the equation: 
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$$ then the direction vector $\{l,m\}$ of the asymptotes are found from the following equation: $$Al^2+2Blm+Cm^2=0$$ (Actually, I don't know the proof) Then to solve this, we let $k=\frac{l}{m}, \ \  (m \not =0)$ and solve the quadratic equation for $k$: $$Ak^2+2Bk+C=0, \text{ in our case } 3k^2+2k-1 = 0$$
From here, I got $k=-1 \text{ or } k=\frac{1}{3}$ (which give us slopes of the two asymptotes). 
Hence we search for the asymptotes of the form $y=kx+b$ and restrict $b$ in such way so that the line does not intersect the line. Plugging this $y$ and $k=-1$ into the equation of the parabola I got: $$(4b-2)x-b^2+10b+14=0$$ 
so $b=\frac{1}{2}$ (as the equation should not have a solution). Then, $y=-x+\frac{1}{2}$ or $2x+2y-1=0$ as in the answer!
However, I could not find the second one in this way...
Then, I got stuck... 
It would be greatly appreciated either if you help me understand (why the asymptotes are to be found so) and complete this solution or suggest another solution.

To justify the approach, note that asymptotically the quadratic terms of the curve 
$f(x,y)=3x^2+2xy-y^2+8x+10y+14$ dominate, i.e.
$$3x^2+2xy-y^2=(3x-y)(x+y)=0$$
which corresponds to the asymptotic behaviors of the asymptotes and yields their slopes $3,\> -1$.
To obtain the actual equations of the two asymptotes, let $f’_x= f’_y=0$ to determine the center, i.e.
$$ 6x+2y+8=0,\>\>\>\>\>2x-2y+10=0$$
Solve to get the center $(-\frac94, \frac{11}4)$. Then, use the point-slope formula for the equations
$$y-\frac{11}4=-(x+\frac94),\>\>\>\>\> y-\frac{11}4=3(x+\frac94)$$
