Question 99620
Is is a standard quadratic eaquation of {{{ax^2+bx+c=0}}} , and can be solved with formula with coefficients: {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} I'll use my favourite notation :-),let's denote s for distance, t - time,a - acceleration,v0-initial velocity. then we have {{{s = (1/2)at^2+v0t}}}
{{{(1/2)at^2+v0t-s=0}}} substituting into formula we get {{{x = (-v0 +- sqrt( v0^2-2*a*s ))/(a) }}} with values we get - getting two answers - -0.0015 and  223.2669 we can forget the negative one because it is phsically imposibble (time can never be negative) - btw those numbers in your equation seem really suspicious to me negative distance....that's either new theory of spacetime or....
anyway quick rememberance:

s...distance[m],distance you travelled
v...velocity[m/s],change of s during infinitesimal change of t ds/dt to be precise
a...acceleration[m/s^2],dv/dt

How to have something from nothing...
now, what I want to show is not fully rigorous, but it is a lot of fun :-),
let's start with a = a with help of our formulae we can state that v=S[t]dt (primitive function) thus v = a*t+c now after deep thought we may state that constant will probably be v0, so v = a*t+v0, followed by s= V[t]dt
s = 1/2a*t^2+v0*t+s0 (thought), now you try to make similiar conclusions, starting with a = 0. Good luck :-D