Real-valued solutions of exponential equation

Let us analyze the Left Hand Side (LHS) and Right Hand Side (RHS):

1. LHS: $2^x + 2^{-x}$. By AM-GM inequality, $2^x + 2^{-x} \ge 2\sqrt{2^x \cdot 2^{-x}} = 2$. Equality holds iff $2^x = 2^{-x} \implies x = 0$.

2. RHS: $2 - (x - 2)^2$. Since $(x - 2)^2 \ge 0$, we have $2 - (x - 2)^2 \le 2$. Equality holds iff $x = 2$.

For LHS to equal RHS, both must equal 2 simultaneously. However, LHS $= 2$ at $x = 0$, while RHS $= 2$ at $x = 2$.

At $x = 0$: LHS $= 2$, RHS $= 2 - 4 = -2$. At $x = 2$: LHS $= 4.25$, RHS $= 2$.

Since there is no value of $x$ where LHS and RHS are equal, the number of real solutions is 0.