jmc

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 $\boxed{8,-112}.$