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(F)=0.8F^2-3
We move all terms to the left:
(F)-(0.8F^2-3)=0
We get rid of parentheses
-0.8F^2+F+3=0
a = -0.8; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-0.8)·3
Δ = 10.6
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{10.6}}{2*-0.8}=\frac{-1-\sqrt{10.6}}{-1.6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{10.6}}{2*-0.8}=\frac{-1+\sqrt{10.6}}{-1.6} $
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