F(n)=n2+2n+5

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Solution for F(n)=n2+2n+5 equation: 



(F)=F2+2F+5

We move all terms to the left:

(F)-(F2+2F+5)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-1F-5=0

a = -1; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·(-1)·(-5)
Δ = -19
Delta is less than zero, so there is no solution for the equation

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