Bernstein Inequality says that if $X_1,X_2,\dots$ are random variables and let $S = \sum_i^n X_i - EX_i$ and if $$\sum_{i=1}^n \mathbb{E}[(X_i)^{q}_{+}] \leq \frac{q!}{2} vc^{q-2}$$
then $$P\left[S \geq \epsilon\right] \leq \exp { - \frac{\epsilon^2}{2(v+c\epsilon/3)}}$$
how do we get inequality of form $$P\left[S \geq F(t) \right] \leq e^{-t} $$ ? for some function $F(t)$. I want to parametrize probability instead of deviation. I am struck here
$$\frac{\epsilon^2}{2(v + c \epsilon/3)} = t \implies \epsilon^2 = t.2.(v+c \epsilon/3)$$
book just concludes that it would be $P[S \geq \sqrt{2vt} + ct] \leq e^{-t}$. I am not sure how do we get this.