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 

Consider the first two equations, under this slightly different form:$$\left\{\begin{array}{l}(1+4\lambda)x+6y=0\\6x+(2+\lambda)y=0.\end{array}\right.$$Suppose that this homogeneous system has exatly one solution; then this solution is $(x,y)=(0,0)$, which is not a solution of the third one.
But\begin{align}\text{The system has more than one solution}&\iff\det\begin{bmatrix}1+4\lambda&6\\6&2+\lambda\end{bmatrix}=0\\&\iff\lambda=-\frac{17}4\vee\lambda=2.\end{align}So, deal only with the cases $\lambda=-\frac{17}4$ and $\lambda=2$. Can you do that?
