F(X)=5-4x+x2

Solution for F(X)=5-4x+x2 equation:

(F)=5-4F+F2

We move all terms to the left:

(F)-(5-4F+F2)=0

We add all the numbers together, and all the variables

-(-4F+F^2+5)+F=0

We get rid of parentheses

-F^2+4F+F-5=0

We add all the numbers together, and all the variables

-1F^2+5F-5=0

a = -1; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·(-1)·(-5)
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{5}}{2*-1}=\frac{-5-\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*-1}=\frac{-5+\sqrt{5}}{-2}
$