303. Solve the equation:

$$
\left(a-x^{2}\right)^{2}=a+x
$$
$\triangle$ Let's expand the brackets, collect all terms on the left side of the equation:

$$
x^{4}-2 a x^{2}-x+a^{2}-a=0
$$

Let's try the following approach: consider this equation as a quadratic in terms of $a$:

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

Find its discriminant:

$$
D=\left(2 x^{2}+1\right)^{2}-4\left(x^{4}-x\right)=4 x^{2}+4 x+1=(2 x+1)^{2}
$$

Now we get:

$$
a=\frac{2 x^{2}+1 \pm(2 x+1)}{2} ; \quad a_{1}=x^{2}+x+1, \quad a_{2}=x^{2}-x
$$

But we need to express $x$ in terms of $a$, not $a$ in terms of $x$. Let's continue:

$$
\begin{aligned}
& x^{2}+x+1-a=0 ; \quad x=\frac{-1 \pm \sqrt{4 a-3}}{2}\left(a \geq \frac{3}{4}\right) \\
& x^{2}-x-a=0 ; \quad x=\frac{1 \pm \sqrt{4 a+1}}{2}\left(a \geq-\frac{1}{4}\right)
\end{aligned}
$$

Answer: if $a \geq 3 / 4$, then

$$
x_{1,2}=(-1 \pm \sqrt{4 a-3}) / 2, \quad x_{3,4}=(1 \pm \sqrt{4 a+1}) / 2
$$

if $-1 / 4 \leq a<3 / 4$, then $x_{1,2}=(1 \pm \sqrt{4 a+1}) / 2$; if $a<-1 / 4$, then there are no solutions.