Numbers of solutions $e^x=-x^2+2x+5 $ Prove that $e^x=-x^2+2x+5 $ have exactly two solutions.
Is it enoguht that Vertex of the parabola is over $y=e^x$ and arms of it looks down

This is a little out of the blue, but...
Since $e^x > 0$ for all $x \in \mathbb{R}$, there can only be two points where RH intersects $e^x$ since $e^x = -x^2+2x+5 \implies x^2-2 x+e^x = 5$.
If we rewrite $x^2-2 x+e^x = 5$ as $x^2-2 x+e^x = y$, observe that its minima is at $x^2-2 x+e^x = W(\frac{e}{2}) (W(\frac{e}{2})+2)-1,\, x = 1-W(\frac{e}{2})$
where $W(n)$ denotes the Lambert W function.
As $x$ tends from $1-W(\frac{e}{2}) \to \infty$ or $1-W(\frac{e}{2}) \to -\infty$, $y\to\infty$, thus it is easy to see that $e^x = -x^2+2x+5$ has only two solutions.
