# t distribution : formula for the degrees of freedom

I understood why we are using a t distribution in this case , because the sample isn't big enough to approximate the true standard deviation of the population by the sample's . But what I can't find anywhere is a formula for finding the degrees of freedom for this t distribution , usually the cases we have is when the STD of X bar is simply S of the sample / radical (n) . In that case the degrees of freedom is n-1 . My intuition suggest that here the degrees of freedom is probably radical (m^2+n^2) Any help

$df\approx\frac{\left(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}\right)^2}{\frac{\left(\frac{s^2_1}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s^2_2}{n_2}\right)^2}{n_2-1}}=\frac{\left(\frac{1.68^2}{28}+\frac{1.3^2}{14}\right)^2}{\frac{\left(\frac{1.68^2}{28}\right)^2}{27}+\frac{\left(\frac{1.3^2}{14}\right)^2}{13}}\approx 32.8$
In combination with the floor function it is $df'=\lfloor 32.8 \rfloor=32$