What is the inverse function of $\ x^2+x$? I think the title says it all; I'm looking for the inverse function of $\ x^2+x$, and I have no idea how to do it. I thought maybe you could use the quadratic equation or something. I would be interesting to know.

Solve $y = x^2+x$, that is, solve the quadratic $x^2+x-y = 0$. This gives
$x = \frac{1}{2}(-1 \pm \sqrt{1+4y})$. Consequently there is an inverse iff $y \geq -\frac{1}{4}$.
To address issues raised in comments:
Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^2+x$. Then since $f(-\frac{3}{2}) = f(\frac{1}{2}) = \frac{3}{4}$ is is clear that $f$ is not injective, furthermore, since $f(x) \geq -\frac{1}{4}$ it is clear that $f$ is not surjective, hence an inverse function, as usually defined, does not exist for $f: \mathbb{R} \to \mathbb{R}$.
This gives a few options:
(1) Define the function $\eta: \mathbb{R} \to 2^{\mathbb{R}}$ by $$ \eta(t) = \begin{cases} \{ \frac{1}{2}(-1 \pm \sqrt{1+4t})\}, && t \geq -\frac{1}{4} \\ \emptyset, && \text{otherwise}\end{cases}$$
Then we have $f(x) = y$ iff $x \in \eta(y)$. This was what I meant by the initial answer (with a lot less baggage).
(2) Restrict the domain of the original function and find a left inverse. For example, we could consider $f$ on the domain $[-\frac{1}{2},\infty)$, and the function $\phi :[-\frac{1}{4},\infty) \to \mathbb{R}$ by $\phi(t) = \frac{1}{2}(-1 + \sqrt{1+4t})$. Then we have $\phi \circ f = \text{Id}$ on the domain 
$[-\frac{1}{2},\infty)$.
(3) Find a suitable right inverse and domain. For example, $f\circ \phi = \text{Id}$ on the domain $[-\frac{1}{4},\infty)$.
