F(n)=4n2+2n

Solution for F(n)=4n2+2n equation:

(F)=4F^2+2F

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4F^2-1F=0

a = -4; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-4}=\frac{0}{-8}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-4}=\frac{2}{-8}
=-1/4
$