Inverse of $g(x)=\frac{x^2+x}{2}$

Question

While I am graphing a function $g$ in desmos, I want to make $g(y)$ be a function in $x$.

I doesn't mean $\frac{x^2+x}{2}$, but I want the function $f$ such that the graph $y=f(x)$ is same to the graph $x=f(y)$, i.e. the inverse of $g(x)=\frac{x^2+x}{2}$.

Can it be written out using simple funtion or is there a proof of negation?

Or is there any other way to graph the function (using desmos' simple technique) (I don't mean to write a program)?

What I have tried:

I think this is may be equation $$\frac{[f(x)]^2+[f(x)]}{2}=x$$, i.e. $$[f(x)]^2+[f(x)]=2x$$ If the R.H.S. is something like $x^2-\frac{1}{4}$ or $x-\frac{1}{4}$, it will be much simpler.

But I have no idea how to solve.

Answer 1

$y =\dfrac{x^2+x}{2} = \dfrac 12 \left(x+\dfrac 12 \right)^2 - \dfrac 18$ is a parabola. Its vertex is at the point $\left( -\dfrac 12, -\dfrac 18 \right)$. That means that $y \ge -\dfrac 18$.

There are two solutions if you solve $x^2+x = 2y$ for $x$. They are

$x = \dfrac{-1 + \sqrt{1+8y}}{2}$ and $x = \dfrac{-1 - \sqrt{1+8y}}{2}$

Answer 2

Rewrite the equation as $x=f(a)$, $$a^2+a-2x=0$$ $$a=\frac{-1\pm\sqrt{1+8x}}{2}$$