Question 49749
Well, your first few steps are fine. Remember, when you solve an equation for a specified variable, in this case it's t, you have to write the equation so that the variable t in on one side of the equals sign and everything else is on the other side.  If you have a t on both sides of the equation, the job won't be finished.  Let's pick it up where you got it right so far:

{{{2x = 2vt-at^2}}} Now this begins to look much like a quadratic equation in t, so let's put it into the standard form for quadratic equations: {{{at^2+bt+c = 0}}}

{{{at^2-2vt+2x = 0}}} This is the same equation you had only it's now in the standard form. Now you can solve this quadratic equation using the quadratic formula: {{{t =(-b+-sqrt(b^2-4ac))/2a}}} In your equation: a = a, b = -2v, and c = 2x

{{{t = (-(-2v)+-sqrt((-2v)^2 - 4(a)(2x)))/2(a)}}} Simplify this.
{{{t = (2v+-sqrt(4v^2-8ax))/2a}}}
{{{t = (2v+-sqrt(4(v^2-2ax)))/2a}}}
{{{t = (2v+-2sqrt(v^2-2ax))/2a}}}

As with any quadratic equation, there are two solutions:

{{{t = (v/a)+(sqrt(v^2-2ax))/a}}} and
{{{t = (v/a)-(sqrt(v^2-2ax))/a}}}