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(F)=16-F^2
We move all terms to the left:
(F)-(16-F^2)=0
We get rid of parentheses
F^2+F-16=0
a = 1; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·1·(-16)
Δ = 65
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{65}}{2*1}=\frac{-1-\sqrt{65}}{2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*1}=\frac{-1+\sqrt{65}}{2} $
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