Orthic Configuration

Using Heron's formula, area for the triangle $ABC$ is $84$ .

The altitudes lengths can be calculated using regular formula for the triangle area. The altitude lengths $h_1,h_2,h_3$ for corresponding vertices $A,B,C$ are $12,56/5, 168/13$ . Let $H$ denote orthocenter of triangle $ABC$ , $D,E,F$ feet of altitudes $h_1,h_2,h_3$ .

$AH \times HD = BH \times HE = CH \times HF$ .

$\triangle AHB + \triangle BHC + \triangle AHC = 84$

$(13\times HF)/2 + (14\times HD)/2 + (15\times HE)/2 =84$

$AH+HD=h_1$ , $BH+HE=h_2$ , $CH+HF=h_3$

After solving these equations we can calculate length $CH=\dfrac{39}{4}$

$CH$ is also a diameter of a circumcircle of triangle $CDE$ and its centerpoint is a midpoint of $CH$ . Therefore $x=\boxed{9.75}$ .