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96f^2-104f+24=0
a = 96; b = -104; c = +24;
Δ = b2-4ac
Δ = -1042-4·96·24
Δ = 1600
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{1600}=40$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-40}{2*96}=\frac{64}{192} =1/3 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+40}{2*96}=\frac{144}{192} =3/4 $
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