Let $ABC$ be a triangle. Let $X$ be a point on $\overline{AC}$ such that $\angle BXA = 90^\circ.$ If $\angle ABC = 90^\circ,$ $AX = 15,$ and $CX = 6,$ then what is $BX?$
Nevermind, I figured it out myself, the answer is: 3sqrt(10)
+6 
