F(n)=2-5n^2

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

(F)=2-5F^2

We move all terms to the left:

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

We get rid of parentheses

5F^2+F-2=0

a = 5; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·5·(-2)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{41}}{2*5}=\frac{-1-\sqrt{41}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*5}=\frac{-1+\sqrt{41}}{10}
$