Solve the equation (a − x2)2 = a + x for x in terms of a.

Let $y = a - x^2$.
The original equation $(a - x^2)^2 = a + x$ can be rewritten as $y^2 = a + x$.
Now we have a system of two equations:

$$y = a - x^2 \quad (1)$$ $$y^2 = a + x \quad (2)$$

Subtract equation (1) from equation (2):

$$y^2 - y = (a + x) - (a - x^2)$$ $$y^2 - y = x + x^2$$

Rearrange the terms to one side:

$$y^2 - x^2 - y - x = 0$$

Factor by grouping:

$$(y - x)(y + x) - (y + x) = 0$$ $$(y + x)(y - x - 1) = 0$$

This gives two possible cases:
Case A: $y + x = 0 \implies y = -x$
Case B: $y - x - 1 = 0 \implies y = x + 1$

Substitute $y = -x$ into equation (1):

$$-x = a - x^2$$

Rearrange into a quadratic equation:

$$x^2 - x - a = 0$$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-a)}}{2(1)}$$ $$x = \frac{1 \pm \sqrt{1 + 4a}}{2}$$

For real solutions, $1 + 4a \ge 0 \implies a \ge -\frac{1}{4}$.

Substitute $y = x + 1$ into equation (1):

$$x + 1 = a - x^2$$

Rearrange into a quadratic equation:

$$x^2 + x + 1 - a = 0$$

Use the quadratic formula:

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1 - a)}}{2(1)}$$ $$x = \frac{-1 \pm \sqrt{1 - 4 + 4a}}{2}$$ $$x = \frac{-1 \pm \sqrt{4a - 3}}{2}$$

For real solutions, $4a - 3 \ge 0 \implies a \ge \frac{3}{4}$.

The possible values for $x$ are:

$$\boxed{x = \frac{1 \pm \sqrt{1 + 4a}}{2}, \quad x = \frac{-1 \pm \sqrt{4a - 3}}{2}}$$