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(F)=F2+20F+40
We move all terms to the left:
(F)-(F2+20F+40)=0
We add all the numbers together, and all the variables
-(+F^2+20F+40)+F=0
We get rid of parentheses
-F^2-20F+F-40=0
We add all the numbers together, and all the variables
-1F^2-19F-40=0
a = -1; b = -19; c = -40;
Δ = b2-4ac
Δ = -192-4·(-1)·(-40)
Δ = 201
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{-(-19)-\sqrt{201}}{2*-1}=\frac{19-\sqrt{201}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{201}}{2*-1}=\frac{19+\sqrt{201}}{-2} $
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