Finding velocity in optimization problem Given $s=-16t^2+192t+144$, what is the velocity when $s=0$? 


This is part of a larger optimization problem which I solved, except for this last part. The critical point occurs at $t=6$, so after $t=6$ the position function is decreasing since the slope ($s^{\prime}$) is negative. 

To find the velocity when $s=0$, I tried $s=-16t^2+192t+144=0$, and quadratic formula gives the roots, which are $t=-12.71$ and $t=0.71$. Shouldn't we use the positive root, $t=0.71$, since $t$ must be positive, because $t$ is time and can't be negative? 

The text answer says $-214.72$ but that comes from using the negative root, and I don't understand why negative time $t$ is used. Is $-214.72$ incorrect?

Thanks.
Multiply $-1$ on both sides of the equation and substitute $s=0$ to get $16t^2-192t-144=0$. Then, solve with result from quadratic variables. Let $a=16$, $b=-192$, $c=-144$ and solve with
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
well Dear kmitov
if we solve the quadratic equation we get to 2 roots
12.708 or -0.708

but this answer is not the answer quoted by user 437158
can you help me on this

further velocity and s has a realtion

ds/dt (differentail) 

if we find velocity by differentiating

it is idependent of variable  s=0

and ds/dt = -32t+192

if we put velocity v = o  then we get the time when body was supposed to be staionery

at t= 192/32=0.1875 second , the body was sationery but where it was from its origin s=o 

it means that we should find instantous velocity when t= 0.1875

so we may say that instanous velocity was vt= 192-6 units when s=0

thanks
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