Answer
$f^{-1}(x)=-\frac{1}{2}+\sqrt{x+\frac{1}{4}}$
Domain: $[-\frac{1}{4},\infty)$
Work Step by Step
$f(x)=x^2+x$, where $x\geq -\frac{1}{2}$ or $x+\frac{1}{2}\geq 0$
$y=x^2+x$
$x^2+x=y$ (Add $1/4$)
$x^2+x+\frac{1}{4}=y+\frac{1}{4}$
$(x+\frac{1}{2})^2=y+\frac{1}{4}$ (Take the square root)
$x+\frac{1}{2}=\sqrt{y+\frac{1}{4}}$
$x=-\frac{1}{2}+\sqrt{y+\frac{1}{4}}$ (Replace $x$ by $f^{-1}(x)$ and $y$ by $x$)
$f^{-1}(x)=-\frac{1}{2}+\sqrt{x+\frac{1}{4}}$
The domain of $f^{-1}$ is
$D_{f^{-1}}=\{x|x+\frac{1}{4}\geq 0\}$
$D_{f^{-1}}=\{x|x\geq -\frac{1}{4}\}$
$D_{f^{-1}}=[-\frac{1}{4},\infty)$
