F(x+1)=4x^2+5x+6

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Solution for F(x+1)=4x^2+5x+6 equation: 



(F+1)=4F^2+5F+6

We move all terms to the left:

(F+1)-(4F^2+5F+6)=0

We get rid of parentheses

-4F^2+F-5F+1-6=0

We add all the numbers together, and all the variables

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

a = -4; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·(-4)·(-5)
Δ = -64
Delta is less than zero, so there is no solution for the equation

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