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(F)=-2F^2-5
We move all terms to the left:
(F)-(-2F^2-5)=0
We get rid of parentheses
2F^2+F+5=0
a = 2; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·2·5
Δ = -39
Delta is less than zero, so there is no solution for the equation
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