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(F)=-16F^2+50F+6
We move all terms to the left:
(F)-(-16F^2+50F+6)=0
We get rid of parentheses
16F^2-50F+F-6=0
We add all the numbers together, and all the variables
16F^2-49F-6=0
a = 16; b = -49; c = -6;
Δ = b2-4ac
Δ = -492-4·16·(-6)
Δ = 2785
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{-(-49)-\sqrt{2785}}{2*16}=\frac{49-\sqrt{2785}}{32} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2785}}{2*16}=\frac{49+\sqrt{2785}}{32} $
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