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(F)=2F^2-6
We move all terms to the left:
(F)-(2F^2-6)=0
We get rid of parentheses
-2F^2+F+6=0
a = -2; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-2}=\frac{-8}{-4} =+2 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-2}=\frac{6}{-4} =-1+1/2 $
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