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(F)=1.5-1.5F^2
We move all terms to the left:
(F)-(1.5-1.5F^2)=0
We get rid of parentheses
1.5F^2+F-1.5=0
a = 1.5; b = 1; c = -1.5;
Δ = b2-4ac
Δ = 12-4·1.5·(-1.5)
Δ = 10
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{10}}{2*1.5}=\frac{-1-\sqrt{10}}{3} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{10}}{2*1.5}=\frac{-1+\sqrt{10}}{3} $
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