How to derive △y from this kinematic equation?

[94tank]

So, my professor was able to derive △y = Vi^2/2g from the following equation: Vf^2 = Vi^2 - 2g△y where Vf = 0

I'm trying to do the same but I'm getting confused/stuck, here's the steps I've taken.
Vf^2 = Vi^2 - 2g△y

Substitute Vf:
Vf^2 = 0
0 = Vi^2 -2g△y

Add +2g to both sides:
0 + 2g = Vi^2 △y

Divide by Vi^2:
2g/Vi^2 = △y

Why do I get the reciprocal of Vi^2/2g ?

 

[Deleted member 4993]

So, my professor was able to derive △y = Vi^2/2g from the following equation: Vf^2 = Vi^2 - 2g△y where Vf = 0

I'm trying to do the same but I'm getting confused/stuck, here's the steps I've taken.
Vf^2 = Vi^2 - 2g△y

Substitute Vf:
Vf^2 = 0
0 = Vi^2 -2g△y

Add +2g to both sides:
0 + 2g = Vi^2 △y ..................................This is incorrect

Add +2g△y to both sides:

2g△y = Vi^2

Divide by 2g both sides

△y = Vi^2/(2g)

Divide by Vi^2:
2g/Vi^2 = △y

Why do I get the reciprocal of Vi^2/2g ?

.

 

[94tank]

.

Wow, thank you so much! I was stuck on this for awhile. I can't just add 2g because it's attached to Y correct? Do you have any tips/recommendations that you use to avoid these type of mistakes? I seem to do them often. :|