Simplistic approach.

Let the arc length of the quadratic Bezier segment
$\operatorname{Bez_2}(A,B,C)$
with control points $A,B,C$ be $L$.
The points of the Bezier segment are found as
\begin{align}
P(t)=(P_x(t),P_y(t))
,\\
P_x(t)&=
A_x\,(1-t)^2+2\,B_x(1-t)t+C_x\,t^2
\tag{1}\label{1}
,\\
P_y(t)&=
A_y\,(1-t)^2+2\,B_y(1-t)t+C_y\,t^2
\tag{2}\label{2}
.
\end{align}
Given $A,C$ and $B_x$,
we can find $t=t_x$ from \eqref{1},
and \eqref{2} will provide the relation
between the $y$-coordinate of the point on the segment
with given $x=B_x$, and corresponding value
of the $y$-coordinate of the control point.
Consider an ellipse with the foci at $A,C$,
which intersects the line $x=B_x$ at the points $E$ and $F$,
\begin{align}
|AE|+|CE|&=|AF|+|CF|=L
.
\end{align}
These points define the range $B_{\max}B_{\min}$
on the line $x=B_x$, where the control point $B$
must be located in order to get the arc length $L$
of the Bezier segment.
So, we can choose any point
\begin{align}
B_t&=B_{\max}(1-t)+B_{\min}\,t
,\quad t\in[0,1]
\end{align}
in that range,
find corresponding control point $B_{\mathrm{mid}}$
and check if the segment
$\operatorname{Bez_2}(A,B_{\mathrm{mid}},C)$
has the length $L$ withing a suitable error range.
In general, there are two control points, $B_1$ and $B_2$,
separated by the line through $A,C$,
that produce two different quadratic Bezier segments
with the same arc length $L$.