F(4y)=2(4y)^2+3

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Solution for F(4y)=2(4y)^2+3 equation: 



(4F)=2(4F)^2+3

We move all terms to the left:

(4F)-(2(4F)^2+3)=0

We get rid of parentheses

-24F^2+4F-3=0

a = -24; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·(-24)·(-3)
Δ = -272
Delta is less than zero, so there is no solution for the equation

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