\(2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \cdots + 2x + 3 = 0\\ (2x + 3)(x^{2020} + x^{2018} + \cdots + 1) = 0\)
Because \(x^{2020} + x^{2018} + \cdots + 1 > 0\), the only possibility is \(2x + 3 = 0\).
Solving gives \(x = -\dfrac32\).
There is only 1 real root.