F(0)=3n^2+4n

Solution for F(0)=3n^2+4n equation:

(0)=3F^2+4F

We move all terms to the left:

(0)-(3F^2+4F)=0

We add all the numbers together, and all the variables

-(3F^2+4F)=0

We get rid of parentheses

-3F^2-4F=0

a = -3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-3)·0
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{-(-4)-4}{2*-3}=\frac{0}{-6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-3}=\frac{8}{-6}
=-1+1/3
$