The function $f(x) = 2e^x + 3x^2 - 2x + 5$ is naturally thought of as a sum of two functions $f_1(x) = 2 e^x$ and $f_2(x) = 3x^2 - 2x + 5$. If we look at $f_1$ and $f_2$ separately, we see that they fail to be surjective "for the same reason": there exist real numbers $M_1$ and $M_2$ such that $f_1(x) \geq M_1$ for all $x$ and $f_2(x) \geq M_2$ for all $x$.
In the case of $f_1$, it is clear that we may take $M_1 = 0$. In the case of $f_2$, this is a well-known property of quadratic functions with positive leading coefficient, which we could prove by completing the square. But we could also prove it in other ways: e.g. $f_2$ is a differentiable function with a unique critical point, which is a minimum: f_2'(x) = 6x - 2, so the critical point is at $x_0 = \frac{1}{3}$ and it has f_2'(x) < 0 for $x < x_0$ and f_2'(x) > 0 for $x > x_0$. Therefore it has a global minimum at $x_0$. Or, following the idea of the OP (which is a good idea), $f_2$ is a continuous function approaching $\infty$ as $x \rightarrow \pm \infty$, so it must have a global minimum.
Finally, of course, if $f(x) = f_1(x)$ is such that $f_1(x) \geq M_1$ for all $x$ and $f_2(x) \geq M_2$ for all $x$, then $f(x) \geq M_1 + M_2$ for all $x$, so $f$ is not surjective.
The preceding considerations are clearly applicable to other functions as well. The text solution, which consists of pure algebraic manipulation, lacks a clear moral and as such seems in danger of being quickly forgotten.