Answer
$$t_{1}=98 \mathrm{s} $$
Work Step by Step
Using $\theta_{1}=\omega_{0} t_{1}+\frac{1}{2} \alpha t_{1}^{2}(\text { Eq. } 10-13)$ and the quadratic formula, we have
$$t_{1}=\frac{-\omega_{0} \pm \sqrt{\omega_{0}^{2}+2 \theta_{1} \alpha}}{\alpha}=\frac{-(0.239 \mathrm{rev} / \mathrm{s}) \pm \sqrt{(0.239 \mathrm{rev} / \mathrm{s})^{2}+2(20 \mathrm{rev})\left(-7.12 \times 10^{-4} \mathrm{rev} / \mathrm{s}^{2}\right)}}{-7.12 \times 10^{-4} \mathrm{rev} / \mathrm{s}^{2}}$$
which yields two positive roots: $98\ \mathrm{s}$ and $572\ \mathrm{s}$ .
Since the question makes sense only if $$t_{1} \lt t_{2}$$ we conclude the correct result is $$t_{1}=98 \mathrm{s} $$
