F(n)=5n2+4n

Solution for F(n)=5n2+4n equation:

(F)=5F^2+4F

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5F^2-3F=0

a = -5; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-5)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-5}=\frac{0}{-10}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-5}=\frac{6}{-10}
=-3/5
$