F(3)=5n^2+2n

Solution for F(3)=5n^2+2n equation:

(3)=5F^2+2F

We move all terms to the left:

(3)-(5F^2+2F)=0

We get rid of parentheses

-5F^2-2F+3=0

a = -5; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-5)·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*-5}=\frac{-6}{-10}
=3/5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*-5}=\frac{10}{-10}
=-1
$