Strange factor in position time function under constant acceleration

Alternate wars says that the time taken for an object to travel a distance under constant acceleration is given by this equation:

$time = 2\sqrt{\frac{distance}{acceleration}}$

For my purposes, I define two points, $\vec{x_{start}}$ and $\vec{x_{end}}$ and calculate the time taken $t$ to travel between them under constant acceleration, $a$ with

$t = 2\sqrt{\frac{|\vec{x_{end}}-\vec{x_{start}}|}{a}}$

Because I have a constant acceleration, Brilliant (and integration...) says that I can find some position at some time, $\vec{x}(t)$ between $\vec{x_{start}}$ and $\vec{x_{end}}$ with the formula

$x-x_0 = v_0t + \frac{1}{2}at^2$

For my purposes, $v_0t$ is zero.

When I substitute $\vec{x}(t)$ for $x-x_0$, and shove the time calculated from $t = 2\sqrt{\frac{|\vec{x_{end}}-\vec{x_{start}}|}{a}}$ into the function, I do not get $\vec{x_{end}}$, rather I get ~$1.9208\vec{x_{end}}$ (a quick google search doesn't bring up anything for this value)

This is true for acceleration values tested between 1 and 500 (steps of 50), and many endpoints.

My current solution is to simply divide by this value and move on, but I'm still curious as to why this is the case.