The equation $$2^x - x^2 = 0$$ has

Compare $$f(x) = 2^x$$ and $$g(x) = x^2$$:

Obvious roots: At $$x = 2$$: $$2^2 = 4 = 2^2$$ ✓. At $$x = 4$$: $$2^4 = 16 = 4^2$$ ✓.

Between $$x = 2$$ and $$x = 4$$: at $$x = 3$$, $$f = 8$$ and $$g = 9$$, so $$g > f$$ — but at both endpoints they're equal, so this gives no new root.

For $$x > 4$$: $$f$$ (exponential) grows faster than $$g$$ (polynomial), so $$f > g$$ forever — no new root.

For $$0 \leq x < 2$$: at $$x = 0$$, $$f(0) = 1$$ and $$g(0) = 0$$, so $$f > g$$. By continuity and $$f(2)=g(2)$$, $$f > g$$ throughout $$[0, 2)$$ — no root here.

For $$x < 0$$: as $$x \to -\infty$$, $$f(x) = 2^x \to 0^+$$ while $$g(x) = x^2 \to +\infty$$, so $$g > f$$. At $$x = 0$$, $$f(0) = 1 > 0 = g(0)$$, so $$f > g$$. Since $$f - g$$ flips sign exactly once on $$(-\infty, 0)$$ (both functions are monotonic on this interval — $$f$$ increasing, $$g$$ decreasing), there is exactly one root, at $$x \approx -0.77$$.

Total: $$\mathbf{3}$$ real solutions ($$x \approx -0.77$$, $$x = 2$$, $$x = 4$$).