Correlation between the equilibrium constant and delta G
The first step is simple:  

$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$

thus

$$ - RT\ln {K_{{\text{eq}}}} \approx  - RT \times 2.303 \times \log {K_{{\text{eq}}}}$$

The second step is terrible because $\text{kcal}$ is used and the units are missing:

$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$

$T = 310{\text{ K}}$ (my guess)

$$\begin{aligned}
  \Delta G =  &amp; \; {-} RT \times 2.303 \times \log {K_{{\text{eq}}}} \\ 
   =  &amp; \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K_{{\text{eq}}}} \\ 
   =  &amp; \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K_{{\text{eq}}}} \\ 
\end{aligned} $$