F(5)=x2+2x+1

Solution for F(5)=x2+2x+1 equation:

(5)=F2+2F+1

We move all terms to the left:

(5)-(F2+2F+1)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-1)·4
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{-(-2)-2\sqrt{5}}{2*-1}=\frac{2-2\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*-1}=\frac{2+2\sqrt{5}}{-2}
$