F(4)=2x^2+32

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Solution for F(4)=2x^2+32 equation: 



(4)=2F^2+32

We move all terms to the left:

(4)-(2F^2+32)=0

We get rid of parentheses

-2F^2-32+4=0

We add all the numbers together, and all the variables

-2F^2-28=0

a = -2; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·(-2)·(-28)
Δ = -224
Delta is less than zero, so there is no solution for the equation

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