F(5)=2x^2+4x+1

Solution for F(5)=2x^2+4x+1 equation:

(5)=2F^2+4F+1

We move all terms to the left:

(5)-(2F^2+4F+1)=0

We get rid of parentheses

-2F^2-4F-1+5=0

We add all the numbers together, and all the variables

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

a = -2; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-2)·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*-2}=\frac{4-4\sqrt{3}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*-2}=\frac{4+4\sqrt{3}}{-4}
$