F(6)=x^2-4x+5

Solution for F(6)=x^2-4x+5 equation:

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

We move all terms to the left:

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

We get rid of parentheses

-F^2+4F-5+6=0

We add all the numbers together, and all the variables

-1F^2+4F+1=0

a = -1; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-1)·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{5}}{2*-1}=\frac{-4-2\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{5}}{2*-1}=\frac{-4+2\sqrt{5}}{-2}
$