426. Solve the equation:

$$
\sqrt{4 x-y^{2}}=\sqrt{y+2}+\sqrt{4 x^{2}+y}
$$
$\triangle$ Let's square this equation; we will obtain an equation equivalent to the given one. We will have:

$$
\begin{aligned}
& 4 x-y^{2}=y+2+4 x^{2}+y+2 \sqrt{(y+2)\left(4 x^{2}+y\right)} \\
& \left(4 x^{2}-4 x+1\right)+\left(y^{2}+2 y+1\right)+2 \sqrt{(y+2)\left(4 x^{2}+y\right)}=0 \\
& (2 x-1)^{2}+(y+1)^{2}+2 \sqrt{(y+2)\left(4 x^{2}+y\right)}=0
\end{aligned}
$$

Therefore,

$$
2 x-1=0, \quad y+1=0, \quad(y+2)\left(4 x^{2}+y\right)=0
$$

The last system of equations has a unique solution ( $\left.\frac{1}{2} ;-1\right)$.

Answer: $(1 / 2 ;-1)$.