Step-by-step explanation:
This is the formula for the change in distance (δx) of an object moving with constant acceleration (a) and initial velocity (v), one of the standard Equations of motion. To solve for t first rearrange
12at2+vt−δx=0
then apply the formula for solving a Quadratic equation if our equation is At2+Bt+C=0 then the solution is
t=−B±B2−4AC√2A,
here A=12a, B=v, C=−δx so plugging these in the formula gives
t=−v±✓v2−4(12a)(−δx)√2(12a) (Apply quandriatic equation)
simplifying
t=−v±v2+2aδx√a.
Note how there are two possible solutions one corresponding to the + and one to the -. Sometimes one of these will give a negative time, so you may want to eliminate that. Indeed if δx>0, v>0 and a>0 we only want the positive solution. In other cases you may want to consider both solutions, for instance if we thow a ball up in the air the initial velocity is positive and the acceleration is negative, it will pass a fixed height above your head twice, once on the way up and once on the way down.