$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$ 
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$

I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\dfrac{\sqrt{55}}{3}$
Are these $\pm$ signs correct? My reasoning came for a CAST diagram, can you confirm if this is correct and if not, why?

Using the Weierstrass transformation, the equation rationalizes to
$$50\left(\frac{1-t^2}{1+t^2}\right)^2+5\frac{1-t^2}{1+t^2}=6\left(\frac{2t}{1+t^2}\right)^2$$
and gives, by solving the biquadratic
$$45t^4-100t^2+55=24t^2,$$
$$t^2=\frac 59,\frac{11}5.$$
The tangent follows, by
$$\pm\frac{2\dfrac{\sqrt5}3}{1-\dfrac 59},\pm\frac{2\sqrt{\dfrac{11}5}}{1-\dfrac{11}5}.$$
Here you don't have to worry about angles nor quadrants.
