Find all real values of $a$ for which the quadratic equation
\[(3 - i) x^2 + (a + 4i) x - 115 + 5i = 0\]has at least one real root.

Enter all the possible values of $a,$ separated by commas.
Let $r$ be the real root.  Then
\[(3 - i) r^2 + (a + 4i) r - 115 + 5i = 0.\]We can write this as
\[(3r^2 + ar - 115) + (-r^2 + 4r + 5)i = 0.\]The real and imaginary parts must both be 0, so $3r^2 + ar - 115 = 0$ and $-r^2 + 4r + 5 = 0.$

The equation $-r^2 + 4r + 5 = 0$ factors as $-(r - 5)(r + 1) = 0,$ so $r = 5$ or $r = -1.$

If $r = 5,$ then
\[3 \cdot 25 + 5a - 115 = 0.\]Solving for $a,$ we find $a = 8.$

If $r = -1,$ then
\[3 \cdot (-1)^2 - a - 115 = 0.\]Solving for $a,$ we find $a = -112.$

Thus, the possible values of $a$ are $\boxed{8,-112}.$