Answer :
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 :)
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 :)