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(-3)=2F^2-5F+3
We move all terms to the left:
(-3)-(2F^2-5F+3)=0
We add all the numbers together, and all the variables
-(2F^2-5F+3)-3=0
We get rid of parentheses
-2F^2+5F-3-3=0
We add all the numbers together, and all the variables
-2F^2+5F-6=0
a = -2; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·(-2)·(-6)
Δ = -23
Delta is less than zero, so there is no solution for the equation
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