How does one-fourth distance in last second give (t-1s)=±√(3/4)?

[mateomy]

Reading a step-by-step problem in my Kinematics chapter; the problem states: "An object is dropped travels one-fourth its distance in the last second of the fall. What height was it dropped from?

They initially use the motion equation:

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

(sans the v(Initial) because it is assumed it is dropped from rest)

I can follow that they have to at first solve for time "t" but they move on down the line of variable rearrangement and come to a certain spot where
I HAVE ABSOLUTELY NO IDEA how they go from this...

$$\frac{1}{4}t^2 = t^2 - (t - 1s)^2$$

to this...

$$(t - 1s) = \pm \sqrt{\frac{3}{4}}$$

I understand that this might be sort of confusing with no real frame of reference within the problem. I was being to lazy to LaTex all of the steps. I can supply more if requested (ugh).
I've been ruminating over this for the last god-knows-how-long. Maybe I am just fried.
Thanks.

 

[tiny-tim]

hi mateomy!

(have a square-root: √ and try using the X² icon just above the Reply box )

that should be (t - 1s) = ±t√(3/4)

 

[mateomy]

Thank you so much, you have no idea how much time I wasted staring at this.