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(F)=2F^2+6F+5
We move all terms to the left:
(F)-(2F^2+6F+5)=0
We get rid of parentheses
-2F^2+F-6F-5=0
We add all the numbers together, and all the variables
-2F^2-5F-5=0
a = -2; b = -5; c = -5;
Δ = b2-4ac
Δ = -52-4·(-2)·(-5)
Δ = -15
Delta is less than zero, so there is no solution for the equation
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