Recall that "Cardano's Formula" gave solutions to the equation $\displaystyle {x}^{{3}}+{m}{x}={n}$ as:

$\displaystyle {x}={\sqrt[{{3}}]{{\frac{{n}}{{2}}+\sqrt{{{R}}}}}}+{\sqrt[{{3}}]{{\frac{{n}}{{2}}-\sqrt{{{R}}}}}}$

where $\displaystyle {R}=\frac{{n}^{{2}}}{{4}}+\frac{{m}^{{3}}}{{27}}$

For the cubic equation, $\displaystyle {x}^{{3}}+{4}{x}={3}$, find the value of $\displaystyle {R}.$

Answer: $\displaystyle {R}=$
Round your answer to two decimal places.