1. We need the number of points where the curve $y=x^5-20x^3+50x+2$ crosses the $x$-axis.

   This means we need the number of real roots of $f(x)=x^5-20x^3+50x+2=0.$

2. To determine how many times the graph can cross the $x$-axis, first study the monotonicity using the derivative.

   $f'(x)=5x^4-60x^2+50=5(x^4-12x^2+10).$

   Let $u=x^2$. Then $x^4-12x^2+10=u^2-12u+10=0.$

   Solving, $u=6\pm\sqrt{26}.$

   Hence the critical points are at $x=\pm\sqrt{6-\sqrt{26}},\quad \pm\sqrt{6+\sqrt{26}}.$

3. Let $a=\sqrt{6-\sqrt{26}},\qquad b=\sqrt{6+\sqrt{26}}.$ Numerically,

   \qquad b\approx 3.33.$$ Since $$f'(x)=5(x^2-(6-\sqrt{26}))(x^2-(6+\sqrt{26})),$$ the sign of $f'(x)$ is: - positive for $|x|<a$ - negative for $a<|x|<b$ - positive for $|x|>b$ Therefore the function behaves as: - increasing on $(-\infty,-b)$ - decreasing on $(-b,-a)$ - increasing on $(-a,a)$ - decreasing on $(a,b)$ - increasing on $(b,\infty)$

4. Now evaluate $f(x)$ at convenient points around these intervals to track sign changes.

   Since $f$ is continuous, each sign change gives a root.

   Compute: $f(-5)=(-5)^5-20(-5)^3+50(-5)+2=-3125+2500-250+2=-873<0$ $f(-4)=(-4)^5-20(-4)^3+50(-4)+2=-1024+1280-200+2=58>0$

   So there is one root in $(-5,-4)$.

   Next, $f(-2)=(-32)-20(-8)-100+2=-32+160-100+2=30>0$ $f(-1)=-1+20-50+2=-29<0$

   So there is one root in $(-2,-1)$.

   Next, $f(0)=2>0$ so between $-1$ and $0$, there is one root in $(-1,0)$.

   Next, $f(1)=1-20+50+2=33>0$ $f(2)=32-160+100+2=-26<0$

   So there is one root in $(1,2)$.

   Next, $f(4)=1024-1280+200+2=-54<0$ $f(5)=3125-2500+250+2=877>0$

   So there is one root in $(4,5)$.

5. Thus we have found 5 distinct real roots, in the intervals:

   \;(-2,-1), \;(-1,0), \;(1,2), \;(4,5).$$

6. Since a quintic polynomial can have at most 5 real roots, these are all the roots.

Therefore, the number of points where the curve crosses the $x$-axis is $\boxed{5}.$