Isolating t in horizontal motion equation

[mom2maxncoop]

Isolating "t" in horizontal motion equation

Homework Statement

So this is just a practice question from my course and they gave me the answer, the problem I am having is understanding how they got to that answer. I have the formula and everything, just don't know how to isolate "t" and the steps to do so


-How long does it take for a flowerpot to fall from a balcony that is 18m above the ground?

Homework Equations

d=½a$\nabla$t²

The Attempt at a Solution

18m [down] = ½(9.8m/s²)$\nabla$t²
and I know I have to isolate the "t" but how do I go about doing that, I keep getting it wrong

The answer is 1.9s,

 

[SteamKing]

Show us what you have done so far.

 

[mom2maxncoop]

After posting this, I sat for another 2 hrs trying to figure out the problem and I finally understood it.

18m [down] = ½(9.8m/s²)∇t²
18(2)=9.8(t²)
36=9.8(t²)
36/9.8 = t²
3.67=t²
squareroot 3.67 = t
1.91=t