F(x)=2x^2+4x+8

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Solution for F(x)=2x^2+4x+8 equation: 



(F)=2F^2+4F+8

We move all terms to the left:

(F)-(2F^2+4F+8)=0

We get rid of parentheses

-2F^2+F-4F-8=0

We add all the numbers together, and all the variables

-2F^2-3F-8=0

a = -2; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·(-2)·(-8)
Δ = -55
Delta is less than zero, so there is no solution for the equation

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