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Solve for tS=9t^2-vt

User Phillippa
by
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1 Answer

4 votes

ANSWER:


t=\frac{v+\sqrt[]{v^2+36S}}{18},\frac{v-\sqrt[]{v^2+36S}}{18}

Explanation:

We have the following equation:


S=9t^2-vt\:

We solve for t:


9t^2-vt-S=0

We can solve by means of the general equation of the quadratic formulas.

Knowing that:

a = 9

b = -v

c = -S


\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}

User LaTeX
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