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(F)=50-3F^2
We move all terms to the left:
(F)-(50-3F^2)=0
We get rid of parentheses
3F^2+F-50=0
a = 3; b = 1; c = -50;
Δ = b2-4ac
Δ = 12-4·3·(-50)
Δ = 601
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{601}}{2*3}=\frac{-1-\sqrt{601}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{601}}{2*3}=\frac{-1+\sqrt{601}}{6} $
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