Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is 
Finding Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is 

what I tried:
let $S = 4x^2+4xy+4y^2+x-5$
$\dfrac{dS}{dx}=8x+4y+1$ and $\dfrac{dS}{dy}=4x+8y$
for center $\dfrac{dS}{dx}=0$ and $\dfrac{dS}{dy}=0$
getting center as $ x=-\dfrac{1}{6}$ and $y=\dfrac{1}{12}$
How do I solve it? Help me please

We have to turn axes through $\theta$ where
$$ \tan 2 \theta = \frac {h}{b-a}=\frac {2}{4-4}= \infty; \theta = \pi/4$$
in order to remove $xy$ term and have axes parallel to $x-,y-,$ axes
$$ x1=(x-y)/\sqrt2;\, y1=(x+y)/\sqrt2;$$
plug in and taking only quadratic terms as $(x,y)$ translations do not affect eccentricity $e,$
$$ \frac12x_1^2+\frac32y_1^2 + ... = 0 $$
$$ e^2=1-\frac{b^2}{a^2} = 1- \frac{2/3}{2}\rightarrow  e=\sqrt{\frac{2}{3}}$$
