First, we need to guarantee $x^2-3x+2\ge 0$. This leads to $$x \ge 2\quad\text{or}\quad x \le 1$$ In fact, this is the domain of the given function. While $x\ge 2$, $x+\sqrt{x^2-3x+2}$ is a monotonically increasing function. It is easy to determine that $y\ge 2$. However, when $x\le 1$, it is no longer a monotonic function. Consequently, it is difficult to determine the function's range. Given we know $x$'s range, it is then possible to express $x$ using $y$ and then solve $y$'s range. $$y=x+\sqrt{x^2 -3x+2} \implies x=\frac{y^2-2}{2y-3}$$ Therefore it must hold that $$\frac{y^2-2}{2y-3} \le 1\quad\text{or}\quad \frac{y^2-2}{2y-3}\ge 2$$ Because we have already handle the $2^{nd}$ case, it is sufficient to just solve $$\frac{y^2-2}{2y-3} \le 1$$ Do casework. - If $2y-3 > 0$, then $y^2-2 \le 2y-3 \implies y=1$. - If $2y-3 < 0$, then $y^2 -2 \ge 2y-3 \implies 1\le y < \frac{2}{3}$. Hence, we conclude $y\in [1, 3/2) \cup [2, +\infty)$.