If 3x+4y=26, and {x,y} are positive integers (which looks likely from the square roots in the second expression), then 3x can only be 6 (since 6+20=26) or 18 (since 18+8=26). This gives {x,y} as {2,5}, {6,2}

Substituting:

√(x2+y2-4x+2y+5)+√(x2+y2-20x-10y+125), x=2,y=5, is 14.

√(x2+y2-4x+2y+5)+√(x2+y2-20x-10y+125), x=6,  y=2, is 10, as desired.

Note that (x2+y2-4x+2y+5) = (x-2)^2+(y+1)^2 

Note that (x2+y2-20x-10y+125) = (x-10)^2+(y-5)^2 

Because the expressions in the second equation are square roots, they must be positive overall. But if x is positive and at least 2, then y could be -1. Trying this: √(x2+1-4x-2+5)+√(x2+1-20x+10+125)=10, when x=2. But {x,y} = {2,-1} would not satisfy the first equation.

So {x,y} = {6,2} is the only real solution.