Set the function equal to zero and because the left side will not factor we’ll need to use the quadratic formula to find the roots of the function.

\[18 - 3t - 2{t^2} = 0\] \[t = \frac{{3 \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( { - 2} \right)\left( {18} \right)} }}{{2\left( { - 2} \right)}} = \frac{{3 \pm \sqrt {153} }}{{ - 4}} = \frac{{3 \pm \sqrt {\left( 9 \right)\left( {17} \right)} }}{{ - 4}} = \frac{{3 \pm 3\sqrt {17} }}{{ - 4}} = - \frac{3}{4}\left( {1 \pm \sqrt {17} } \right)\]

So, the quadratic formula gives the following two roots of the function,

\[ - \frac{3}{4}\left( {1 + \sqrt {17} } \right) = - 3.842329\hspace{0.25in}\hspace{0.25in} - \frac{3}{4}\left( {1 - \sqrt {17} } \right) = 2.342329 \]