Respuesta :

kritter1
You would have to replace f(x) with zero and use quadratic formula 

f(x)=x^2-5x+5 
0=x^2-5x+5 

=(-b+-√b^2-4ac)/2a 
=(5+-√25-4(1)(5))/2
=(5+-√5)/2 

x=(5+√5)/2 or x=(5-√5)/2 

Both x values are completely applicable in the equation. 

Hope I helped :)