The number of real roots of the equation,

Let $e^{x}=t \in(0, \infty)$

Given equation

$t^{4}+t^{3}-4 t^{2}+t+1=0$

$\Rightarrow t^{2}+t-4+\frac{1}{t}+\frac{1}{t^{2}}=0$

$\Rightarrow\left(t^{2}+\frac{1}{t^{2}}\right)+\left(t+\frac{1}{t}\right)-4=0$

Let $t+\frac{1}{t}=y$

$\left(y^{2}-2\right)+y-4=0 \Rightarrow y^{2}+y-6=0$

$y^{2}+y-6=0 \Rightarrow y=-3,2$

$\Rightarrow y=2 \quad \Rightarrow \quad t+\frac{1}{t}=2$

$\Rightarrow \quad e^{x}+e^{-x}=2$

$x=0$, is the only solution of the equation

Hence, there only one solution of the given equation.