F(3)=2x2+1

Solution for F(3)=2x2+1 equation:

(3)=2F^2+1

We move all terms to the left:

(3)-(2F^2+1)=0

We get rid of parentheses

-2F^2-1+3=0

We add all the numbers together, and all the variables

-2F^2+2=0

a = -2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-2)·2
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-2}=\frac{-4}{-4}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-2}=\frac{4}{-4}
=-1
$