The position function of a particle is given by.....

The position function of a particle is given by

s=t^3−5t^2−8t, t≥0

where s is measured in meters and t in seconds. When does the particle reach a velocity of 15 m/s?

so i got to v(s) = 3t^2-10t-8
15 = 3t^2-10t-8
0 = 3t^2-10t-23

but not sure how to do from here :S

It is easy to forget algebra once you start working on calculus- in fact, that can sometimes be the hardest part. The answer is quite simple- use the quadratic formula.
If you remember, the quadratic formula says that

x =( -b +- (b^2 + 4 *a*c)^.5 ) / (2*a)

in this case

x =( 10 +- (100 + 276)^.5 )/ 6

this gives two solutions. ( +- means plus or minus) However, one of the solutions is negative, which is impossible because the problem stated that t must be nonnegative.

the answer is then t = (10 + (376)^.5 ) /6

Hope that helps!

Then you solve 0 = 3t^2-10t-23

which leads to t = -1.56 or 4.9

-1.56 does not make sense since you need t≥0;
so the answer is 4.9 sec.