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(F)=-4F^2+6
We move all terms to the left:
(F)-(-4F^2+6)=0
We get rid of parentheses
4F^2+F-6=0
a = 4; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·4·(-6)
Δ = 97
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{97}}{2*4}=\frac{-1-\sqrt{97}}{8} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{97}}{2*4}=\frac{-1+\sqrt{97}}{8} $
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