Solve t'2-2t+300=0

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t - (2 t + 5) - 5 (1 - 2 t) = 2 (3 + 4 t) - 3 (t - 4)

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\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t

\frac{3 t - 2}{3} + \frac{2 t + 3}{2} = t + \frac{7}{6}

\frac{3 t - 2}{4} - \frac{2 t - 2}{3} = \frac{2}{3} - t

\frac{3 t - 2}{3} + \frac{2 t + 3}{2} = t + \frac{7}{6}

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