F(n)=n2+4n+6

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Solution for F(n)=n2+4n+6 equation: 



(F)=F2+4F+6

We move all terms to the left:

(F)-(F2+4F+6)=0

We add all the numbers together, and all the variables

-(+F^2+4F+6)+F=0

We get rid of parentheses

-F^2-4F+F-6=0

We add all the numbers together, and all the variables

-1F^2-3F-6=0

a = -1; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·(-1)·(-6)
Δ = -15
Delta is less than zero, so there is no solution for the equation

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