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(F)=2F^2-2F-12
We move all terms to the left:
(F)-(2F^2-2F-12)=0
We get rid of parentheses
-2F^2+F+2F+12=0
We add all the numbers together, and all the variables
-2F^2+3F+12=0
a = -2; b = 3; c = +12;
Δ = b2-4ac
Δ = 32-4·(-2)·12
Δ = 105
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{-(3)-\sqrt{105}}{2*-2}=\frac{-3-\sqrt{105}}{-4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{105}}{2*-2}=\frac{-3+\sqrt{105}}{-4} $
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