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(F)=6(F)^2-4
We move all terms to the left:
(F)-(6(F)^2-4)=0
We get rid of parentheses
-6F^2+F+4=0
a = -6; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·(-6)·4
Δ = 97
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{97}}{2*-6}=\frac{-1-\sqrt{97}}{-12} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{97}}{2*-6}=\frac{-1+\sqrt{97}}{-12} $
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