The curvature for a parameterized curve $B(t) = ((x(t), y(t))$ is given by [1]
$$ \kappa(t) = \frac{\left|B'(t), B''(t)\right|}{|| B'(t)||^2}, $$
where the numerator is the determinant of the matrix formed by concatenating $B'(t)$ and $B''(t)$.
Note that the curvature is a function of the parameter $t$, the curvature is not necessarily constant over the curve.
A quadratic Bezier curve is defined by the points $P_0$, $P_1$ and $P_2$ is parameterized by [2]
$$ B(t) = \left(1 - t\right)\left[\left(1 - t\right) P_0 + P_1\right]
+ t \left[ \left(1 - t\right) P_1 + t P_2 \right], $$
with derivatives
$$ B'(t) = 2\left(1 - t\right)\left(P_1 - P_0\right) + 2t\left(P_2 - P_1\right) $$
and
$$ B''(t) = 2\left(P_2 - 2P_1 + P_0\right). $$
Substituting these into the expression for the curvature (using the bilinearity of the determinant operator and the fact that $\left|x,x\right|\equiv0$) yields the numerator
$$\begin{align}
n(t)
&= \left|B'(t), B''(t)\right| \\
&= 4(1-t)\left|P_1-P_0, P_0 - 2P_1 + P_2\right| \\
&\quad+ 4t\left|P_2-P_1, P_0 - 2P_1 + P_2\right| \\
&= 4(1-t)\left|P_1-P_0, P_2-P_1\right| + 4t\left|P_2-P_1, P_0-P_1\right| \\
&= 4\left| P_1-P_0, P_2-P_1 \right|.
\end{align}$$
The denominator is given by
$$ m(t) = ||B(t)||^3, $$
with
$$\begin{align}
||B(t)||^2
&= 4(1-t)^2 ||P_1 - P_0||^2
+ 8t(1-t)(P_1 - P_0)\cdot(P_2 - P_1)
+ 4t^2||P_2 - P_1||^2.
\end{align}$$
As I originally came here in search for maximum curvature of a quadratic Bezier curve, I will also present that here, even if it is not strictly in the question.
The maximum curvature is found at either (i) the maximum of the function $\kappa(t)$ or (ii) one of the endpoints of the curve if the maximum lies outside the range $(0,1)$.
The maximum of the function $\kappa(t)$ corresponds to $\kappa'(t) = 0$, i.e.
$$ \kappa'(t) = \frac{n'(t) m(t) - n(t) m'(t)}{m(t)^2}. $$
Given that the numerator $m(t)$ is a constant, finding zeros of $\kappa'(t)$ equates to finding zeros of $m'(t)$, which in turn reduces to finding zeros of $||B'(t)||^2$. This is given by
$$ \frac{\mathrm{d}}{\mathrm{d}t} ||B(t)||^2 =
8(P_1 - P_0) \cdot (P_0 - 2P_1 + P_2) +
8t || P_0 - 2P_1 + P_2 ||, $$
which gives us the optimal parameter value
$$ t^* = \frac{(P_1 - P_0) \cdot (P_0 - 2P_1 + P_2)}{|| P_0 - 2P_1 + P_2 ||}. $$
Substituting this in the expression and some more algebra yields
$$ \kappa(t^*) = \frac{||P_2 - 2P_1 + P_0||}{2|P_1 - P_0, P_2 - P_1|}. $$
Hope this helps (someone, somewhere, somewhat, someday)
[1] https://en.wikipedia.org/wiki/Curvature
[2] https://en.wikipedia.org/wiki/B%C3%A9zier_curve#Quadratic_B%C3%A9zier_curves