TITLE: Number of real solutions to $x^7 + 2x^5 + 3x^3 + 4x = 2018$
QUESTION [5 upvotes]: Find the number of real solutions of $x^7 + 2x^5 + 3x^3 + 4x = 2018$?
What is the general approach to solving this kind of questions? I am interested in the thought process. 

Few of my thoughts after seeing this question:
since $x$ has all odd powers so, it can not have any negative solution. 
2018 is semiprime; not much progress here. We can sketch the curve but graphing a seven order polynomial is difficult.

REPLY [4 votes]: Render
$x^7+0x^6+2x^5+0x^4+3x^3+0x^2+4x-2018=0$
Descartes' Rule of Signs forces exactly one positive root and no negative roots.  That's it!