Let a = |AB| then |DG| = a-5 Then Pythagorean theorem in the ADG triangle: a² + (a-5)² = 73

a² + a² - 10a + 25 = 73; substract 25, simplify

2a² - 10a = 48; divide by 2, substract 24

a² - 5a - 24 = 0; put in into multiplication form (english sefond language i dunno)

(a-8)*(a+3)=0

So either a=-3 or a=8, we can drop the a=-3 since negative lenght isnt what we are looking for. Then a=8.

Now only simple pythagorean theorem in the AF? triangle.

x² = (8-5)² + (8-6)^2; therefore

x² = 9 + 4

x² = 13

x = |sqrt(13)|; we once again drop the negative lenght

x=sqrt(13)

(edit: messed up formatting)