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(F)=3F^2-4F+10
We move all terms to the left:
(F)-(3F^2-4F+10)=0
We get rid of parentheses
-3F^2+F+4F-10=0
We add all the numbers together, and all the variables
-3F^2+5F-10=0
a = -3; b = 5; c = -10;
Δ = b2-4ac
Δ = 52-4·(-3)·(-10)
Δ = -95
Delta is less than zero, so there is no solution for the equation
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