F(2x)=x^2+5x+3

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Solution for F(2x)=x^2+5x+3 equation: 



(2F)=F^2+5F+3

We move all terms to the left:

(2F)-(F^2+5F+3)=0

We get rid of parentheses

-F^2+2F-5F-3=0

We add all the numbers together, and all the variables

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

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

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