F(3)=2x^2+32

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Solution for F(3)=2x^2+32 equation:



(3)=2F^2+32
We move all terms to the left:
(3)-(2F^2+32)=0
We get rid of parentheses
-2F^2-32+3=0
We add all the numbers together, and all the variables
-2F^2-29=0
a = -2; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·(-2)·(-29)
Δ = -232
Delta is less than zero, so there is no solution for the equation

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