F(4x)=x^2+5

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



(4F)=F^2+5

We move all terms to the left:

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

We get rid of parentheses

-F^2+4F-5=0

We add all the numbers together, and all the variables

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

a = -1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·(-1)·(-5)
Δ = -4
Delta is less than zero, so there is no solution for the equation

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