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(F)=2F^2-1
We move all terms to the left:
(F)-(2F^2-1)=0
We get rid of parentheses
-2F^2+F+1=0
a = -2; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-2)·1
Δ = 9
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{9}=3$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*-2}=\frac{-4}{-4} =1 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*-2}=\frac{2}{-4} =-1/2 $
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