Answer
The inverse function is
$$f^{-1}(x)=\frac{1}{2}+\frac{\sqrt{1+4x}}{2}.$$
Work Step by Step
Notice that $y=x^2-x=x(x-1)$ is a quadratic function. Its zeroes are $x_1=0$ and $x_2=1$. Further, we know that it will achieve its minimum in the point in the middle between its zeroes, i.e. when $x=1/2$ and the value of that minimum is $y_m=(1/2)^2-1/2=-1/4$. This means that in the given domain $x\geq1/2$ the function passed its minimum and is only increasing.
Now we have to express $x$ in terms of $y$.
$$y=x^2-x\Rightarrow x^2-x-y=0.$$
This is a quadratic equation whose solutions are given by
$$x=\frac{1\pm\sqrt{1-4\cdot1(-y)}}{2}=\frac{1\pm\sqrt{1+4y}}{2}=\frac{1}{2}\pm\frac{\sqrt{1+4y}}{2}.$$
we have shown that $y\geq1/4$, so for $x$ to be greater than or equal to $1/2$ we must take the '+' sign only i.e.
$$x=\frac{1}{2}+\frac{\sqrt{1+4y}}{2}.$$
This gives
$$f^{-1}(y)=\frac{1}{2}+\frac{\sqrt{1+4y}}{2}.$$
Renaming $y$ back to $x$ this becomes
$$f^{-1}(x)=\frac{1}{2}+\frac{\sqrt{1+4x}}{2}.$$
