How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$? 
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ? 

My trial:
from given equation:  $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
But the term of $xy$ is a bottleneck. 
From here I can't proceed. 
Can somebody please help me solve this problem? 
Thank you. 

The ellipse we have is in the form $S(x, y) = ax^2 + by^2 + 2hxy + c$. To remove the $xy$ term we rotate the ellipse by the angle $\tan (2\theta) = \dfrac{2h}{a - b}$. Which is same as rotating coordinate system by $-\theta$.
So if $(X', Y')$ is coordinates in new coordinate system then we can use
$${x \choose y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}$$ 
to find relation between old coordinate system $(x,y)$ and new coordinate system $(X', Y')$.
Plugging $ \theta = -\pi/4$
$${x \choose y} = \dfrac{1}{\sqrt{2}}\left[\begin{matrix}1 & 1 \\ -1& 1 \end{matrix}\right] {X^\prime \choose Y^\prime} = \dfrac{1}{\sqrt{2}}{X' +Y'\choose Y'-X'}$$
So, $$10x^2+14xy+10y^2-7= 3 X'^2 + 17 Y'^2 - 7  \implies \dfrac{X'^2}{7/3} + \dfrac{Y'^2}{7/17} = 1$$
The major axis is $2\sqrt{\dfrac73}$ and minor axis is $2\sqrt{\dfrac7{17}}$.
