A hemispherical bowl has a radius of 3.7m. Water is added to the bowl at the rate of $\displaystyle pi/3$ $\displaystyle m^3/min$. When the depth of water in the bowl is $\displaystyle h$ m, and volume of water is given by $\displaystyle V= (1/6)pi*h^2(9-2h)m^3$. How deep is the water after 2.5 minutes? At what rate is the depth increasing at this time?

So I'm guessing that the $\displaystyle m^3$ is just a unit and not part of the formula.

The volume of water in the bowl is $\displaystyle pi/3 * 2.5 = 5pi/6

$

So $\displaystyle 5pi/6=pi/6* h^2(9-2h)$

simplify: $\displaystyle 5=9h^2 - 2h^3$

If I'm right so far how do I solve for h?