If you have a 't' term AND a 't^2' term, the the equation is quadratic and that makes the math a bit more painful - probably too painful for Physics 20. There are two scenarios. Scenario 1: d is zero. That helps a lot; we have 0 = v_ft - 1/2at^2, we can factor t out of it to obtain 0 = t(v_f - 1/2at) and that has two straightforward solutions, t=0 and t=2v_f/a. Scenario 2: d is NOT zero. Now the equation is a trinomial quadratic. If you're incredibly lucky it will be factorable, but generally we'd have to use the quadratic formula to find the value(s) of t. The formatting here is so limited, I don't think I can demonstrate the steps so I'll hope you know what I mean.

@@NorQuestCollegeOnlineAnytime I didn't realize at first, but most of the questions with d=vft-1/2a(t^2) that my teacher gives either have vf or vi = 0, or when that is not the case and I have to solve for t, I can use a different kinematic equation to find a new piece of information using my known variables. Then I can find an equation with the original variables plus the new one and solve it for t. I think I understand better now and thanks for taking the time to help.

Nevermind Its the same method as solving for a.