Find the critical numbers of the function. (Round your answers to three decimal places.)
s(t) = 3t4 + 4t3 − 6t2
Set $\displaystyle s'(t)=0$.
You get (I assume by t2 you mean $\displaystyle t^2$ for example) $\displaystyle 12t^3 + 12t^2 - 12t = 0$.
$\displaystyle t=0$ is an obvious solution...
Divide through by $\displaystyle 12t$ to get
$\displaystyle t^2 + t - 1 = 0$
Solve this using standard $\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ formula