I'm trying to rearange this

0 = Xo + 2^-1 a t^2
for t
I've been told that it is

(a^-1 2x)^(2^-1)

but I do not see how please show me step by step how you get that

0 - Xo = Xo - Xo + 2^-1 a t^2
(-Xo = 2^-1 a t^2) a^-1 2
(a^-1 (2-Xo) = t^2)^(2^-1)
(a^-1 (2 - Xo))^(2^-1)

please show me how to do this as i'm getting negetive number imaginary
show me step by step how to rearang please not just the rearanged equation but how you rearange it because i do not get it

see mathmates response to the question before. I don't understand how you make it so complicated.

x=x0+ vo*t +1/2 gt^2
let x0, vo be zero (so x downward will be negative).

x=1/2 g t^2

t=sqrt (2x/g)-8.8 sec about

To rearrange the equation 0 = Xo + (2^-1)at^2 for t, you can follow these steps:

Step 1: Start with the given equation
0 = Xo + (2^-1)at^2

Step 2: Subtract Xo from both sides to isolate the term containing t^2
- Xo = (2^-1)at^2

Step 3: Divide both sides by (2^-1)a to solve for t^2
(- Xo/((2^-1)a)) = t^2

Step 4: Simplify the expression (- Xo/((2^-1)a))
To simplify, we can multiply the numerator and denominator by 2 to eliminate the fraction:
((- Xo * 2)/((2^-1)a * 2)) = t^2
(- 2Xo/(2^0 * a * 2)) = t^2
(- 2Xo/(2^1 * a * 2)) = t^2
(- 2Xo/(2 * a * 2)) = t^2
(- Xo/(a * 2)) = t^2

Step 5: Take the square root of both sides to solve for t
√((- Xo/(a * 2))) = t
(t = (√(- Xo/(a * 2))))

Note: The rearranged equation might yield a negative or imaginary result depending on the values of Xo and a. If you encounter a negative or imaginary value, make sure to check the original equation and the values you've substituted.