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(F+1)=3F^2+6F-6
We move all terms to the left:
(F+1)-(3F^2+6F-6)=0
We get rid of parentheses
-3F^2+F-6F+1+6=0
We add all the numbers together, and all the variables
-3F^2-5F+7=0
a = -3; b = -5; c = +7;
Δ = b2-4ac
Δ = -52-4·(-3)·7
Δ = 109
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{-(-5)-\sqrt{109}}{2*-3}=\frac{5-\sqrt{109}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{109}}{2*-3}=\frac{5+\sqrt{109}}{-6} $
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