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(F)=0.5-2F^2
We move all terms to the left:
(F)-(0.5-2F^2)=0
We get rid of parentheses
2F^2+F-0.5=0
a = 2; b = 1; c = -0.5;
Δ = b2-4ac
Δ = 12-4·2·(-0.5)
Δ = 5
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{5}}{2*2}=\frac{-1-\sqrt{5}}{4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5}}{2*2}=\frac{-1+\sqrt{5}}{4} $
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