The given equation will be true in 3 cases.
Case 1: when x2−5x+5=1
⇒x2−5x+4=0
⇒x=1,4
Case 2: when x2+4x−60=0
⇒x=6,−10
Case 3: when x2−5x+5=−1 and x2+4x−60∈ even integers
Now, x2−5x+5=−1
⇒x=2,3
Only x=2 satisfies the given condition,
Hence, sum of all real values of x is 1+4+6−10+2=3