ANSWER:
[tex]t=\frac{v+\sqrt[]{v^2+36S}}{18},\frac{v-\sqrt[]{v^2+36S}}{18}[/tex]STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]S=9t^2-vt\: [/tex]We solve for t:
[tex]9t^2-vt-S=0[/tex]We can solve by means of the general equation of the quadratic formulas.
Knowing that:
a = 9
b = -v
c = -S
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{ Replacing:} \\ t=\frac{-(-v)\pm\sqrt[]{(-v)^2-4\cdot9\cdot(-S)}}{2\cdot9}=\frac{v\pm\sqrt[]{v^2+36S}}{18} \\ t_1=\frac{v+\sqrt[]{v^2+36S}}{18} \\ t_2=\frac{v-\sqrt[]{v^2+36S}}{18} \end{gathered}[/tex]