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

The eccentricity only depends on the ratio of the axis so that only the quadratic terms matter.
By symmetry, it is obvious that one of the axis is parallel to $x=y$ and we apply the orthogonal transformation
$$x=u-v,y=u+v$$
which yields
$$3u^2+v^2.$$
Hence
$$e=\sqrt{1-\dfrac13}.$$

More generally, when the quadratic terms are $ax^2+2bxy+cy^2$, we compute the ratio of the Eigenvalues of
$$\begin{pmatrix}a&b\\b&c\end{pmatrix}$$ which are the roots of 
$$x^2-(a+c)x+ac-b^2,$$
and that ratio is
$$\dfrac{a+c-\sqrt{(a-c)^2+4b^2}}{a+c+\sqrt{(a-c)^2+4b^2}}.$$
