F(n)=n^2-5

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

(F)=F^2-5

We move all terms to the left:

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

We get rid of parentheses

-F^2+F+5=0

We add all the numbers together, and all the variables

-1F^2+F+5=0

a = -1; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-1)·5
Δ = 21
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{21}}{2*-1}=\frac{-1-\sqrt{21}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21}}{2*-1}=\frac{-1+\sqrt{21}}{-2}
$