Finding $P(X + 10/X > 7 )$ where $X$ is uniform 
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?

Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \left( X + \frac{10}{X} > 7 \right) = P[(X-5)(X-2)>0] = P[ \{X>2 \} \cup \{ X < 5 \} ] $$
Since we have two cases: either $X>5$ and $X>2$ OR $X<5$ and $X<2$. now, using inclusion-exclusion we get
$$ P(X>2) + P(X<5) - P(2<X<5) = 1 - F(2) + F(5) - F(5) + F(2) = 1$$
What is wrong with my solution? According to my solution key, I should get $\frac{7}{10}$. What am I missing here?

Note: $\mathsf P(X+\tfrac {10}X>7) ~{=~ \mathsf P(X^2+10>7X)\\ =~ \mathsf P((X-5)(X-2)>0) \\=~ \mathsf P(X<2~\cup~5<X)}$
Plot $~y=(x-5)(x-2)~$ and look at where it is above the x-axis.
