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(F)=F^2-10F-96
We move all terms to the left:
(F)-(F^2-10F-96)=0
We get rid of parentheses
-F^2+F+10F+96=0
We add all the numbers together, and all the variables
-1F^2+11F+96=0
a = -1; b = 11; c = +96;
Δ = b2-4ac
Δ = 112-4·(-1)·96
Δ = 505
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{-(11)-\sqrt{505}}{2*-1}=\frac{-11-\sqrt{505}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{505}}{2*-1}=\frac{-11+\sqrt{505}}{-2} $
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