solve the Lagrange multiplier equations the equations :
$$\left\{\begin{array}{l}{x+6 y+4 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
I've done many transformations,but I still can't get the answer.
the last I did this:
$$\lambda=\left(6 \frac{x}{y}+2\right)=\left(\frac{1}{4}+\frac{3}{2} \frac{y}{x}\right),y=tx \\=\left(6 \cdot \frac{1}{t}+2\right)=\left(\frac{1}{4}+\frac{3}{2} t\right)\\t^{2}-\frac{7}{6} t-4=0$$
the  quadratic formula's roots have square number , but the answer doesn't have,so I think I might be wrong.
I don't know how to solve it 

From the first equation we get
$$\frac{y}{x}=-\frac{1+4\lambda}{6}$$
and from the second one
$$\frac{y}{x}=-\frac{6}{2+\lambda}$$
So you will get
$$\frac{1+4\lambda}{6}=\frac{6}{2+\lambda}$$
Can you finish?
