The accelration formula is this:

a = (v_f - v_0)/t

You can rewrite it to look like this:

v_f = v_0 + at

v_f = (1.32944 x 10^(-7))t

All you need is JUST the time, you don't need to find the square root of anything.

To prove it can actually be written that way is this:

You have a position-time function x(t):

x(t) = x_0 + v_0t + 1/2at^2

We know that the derivative of x(t) = v(t) or a velocity-time function

The derivative of the constant x_0 = 0

The derivative of v_0t = v_0 because the derivative of t^n = nt^n-1, so if v_0t then 1 * v_0 = 1v_0 = v_0 and t^(1-1) = t^0 = 1: v_0 * 1 = v_0

The derivative of 1/2at^2 = at because the derivative of t^n = nt^n-1, so if 1/2at^2 then 2 * 1/2a = 1a = a and t^(2-1) = t^1 = t: a * t = at

v(t) = v_0 + at

Thus proving our formula is correct where v(t) = v_f

Another way is to take the given acceleration formula:

a = (v_f - v_0)/t

multiply by t on both sides:

at = v_f - v_0

Add v_0 to both sides:

at + v_0 = v_f

This ALSO proves the formula above.

There are no radicals involved in this.