The sum of all real values of x satisfring the equationx2−5x+5x2+4x−60=1 is

Given, x2−5x+5x2+4x−60=1Clearly, this is possible whenI. x2+4x−60=0 and x2−5x+5≠0                            orII. x2−5x+5=1                            orIII.x2−5x+5=−1 and x2+4x−60= Even integerCase l: When x2+4x−60=0⇒ x2+10x−6x−60=0⇒ x(x+10)−6(x+10)=0 ⇒ (x+10)(x−6)=0 ⇒ x=−10 or x=6 Note that, for these two values of x,                            x2−5x+5≠0Case ll: When x2−5x+5=1⇒ x2−5x+4=0 ⇒ x2−4x−x+4=0 ⇒ x(x−4)−1(x−4)=0 ⇒ (x−4)(x−1)=0 ⇒ x=4 or x=1Case III: When x2−5x+5=−1⇒ x2−5x+6=0 ⇒ x2−2x−3x+6=0 ⇒ x(x−2)−3(x−2)=0 ⇒ (x−2)(x−3)=0 ⇒ x=2 or x=3Now, when x=2,x2+4x−60= 4 + 8 - 60 = - 48, which is an even integer.When x=3,x2+4x−60=9+12−60=−39which is not an even integer.Thus, in this case, we get x = 2.Hence, the sum of all real values of             x=−10+6+4+1+2=3