The distance, d (in metres), which an object travels. is given by the equation $\displaystyle d=ut-8t^2$ where u is the initial velocity (in metres per second) and t is the time in seconds it has been travelling. Show that there is only one value for t when $\displaystyle u^2 =32d$

I'm having real trouble solving this equation for t using any algebraic method that I've learnt for quadratic equations.
I can't figure out how to put it into the quadratic formula simply for the fact that i can't get any of the variables by themself on either side of the equation to set as x.

The steps I've tried so far in order to rearrange for t are:
$\displaystyle d=ut-8t^2$
$\displaystyle d+8t^2=ut$
divide both sides by t
$\displaystyle d/t+8t=u$
take d/t away from both sides
$\displaystyle 8t=u-d/t$
and here is where i lose track, because i figure out that i can't rearrange for t at all. i've considered converting d/t into V (final velocity) but i can't see that helping much.
Any suggestions would be much appreciated.

2. Originally Posted by mattty
The distance, d (in metres), which an object travels. is given by the equation $\displaystyle d=ut-8t^2$ where u is the initial velocity (in metres per second) and t is the time in seconds it has been travelling. Show that there is only one value for t when $\displaystyle u^2 =32d$

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Rearrange the expression for $\displaystyle d$:

$\displaystyle 8t^2-ut+d=0$

This is an equation relating $\displaystyle t,\ u$ and $\displaystyle d$ .

If for some particular $\displaystyle d$ and $\displaystyle u$ we wish to solve for $\displaystyle t$ we will use the quadratic formula. Now the discriminant for the quadratic is:

$\displaystyle D=u^2-32d$

and it is well known that we have no real roots for $\displaystyle t$ if $\displaystyle D<0$, two real roots if $\displaystyle D>0$, and exactly one real root if $\displaystyle D=0$.

The last condition above is the one applicable here.

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3. Many thanks for this, apparently maths at 3am isn't my strong point