(x2−5x+5)x2+4x−60=1 Case I x2−5x+5=1 and x2+4x−60 can be any real number ⇒x=1,4 Case II x2−5x+5=−1 and x2+4x−60 has to be an even number ⇒x=2,3 where 3 is rejected because for x=3, x2+4x−60 is odd Case III x2−5x+5 can be any real number and x2+4x−60=0 ⇒x=−10,6 ⇒ Sum of all values of x =−10+6+2+1+4=3