Prove that equation , accepts an unique resolution : and :
put $\displaystyle g(x)=f(x)-x$, then:
$\displaystyle g'(x)=(-2x^2+x+3)e^{-x}-1$
Now look at $\displaystyle -2x^2+x+3$ this has roots at $\displaystyle x=-1$ and $\displaystyle 3/2$, so $\displaystyle -2x^2+x+3$ has constant sign for $\displaystyle x>3/2$, a quick check shows that this is negative. Hence $\displaystyle g'(x)<0$ for $\displaystyle x>3/2$
Thus $\displaystyle g(x)$ is decreasing when $\displaystyle x>3/2$. To complete the proof it remains to show that $\displaystyle g(3/2)$ is positive and $\displaystyle g(2)$ is negative.
CB