Mathematics - Quadratic Equation Question with Solution | TestHub

The given equation will be true in$3$cases.

Case 1:when$x^2-5x+5=1$

$\Rightarrow x^2-5x+4=0$

$\Rightarrow x=1, 4$

Case 2:when$x^2+4x-60=0$

$\Rightarrow x=6, -10$

Case 3:when$x^2-5x+5=-1$and$x^2+4x-60\in$even integers

Now,$x^2-5x+5=-1$

$\Rightarrow x=2, 3$

Only$x=2$satisfies the given condition,

Hence, sum of all real values of$x$is$1+4+6-10+2$$=3$