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(F)=3F^2-36-6
We move all terms to the left:
(F)-(3F^2-36-6)=0
We get rid of parentheses
-3F^2+F+36+6=0
We add all the numbers together, and all the variables
-3F^2+F+42=0
a = -3; b = 1; c = +42;
Δ = b2-4ac
Δ = 12-4·(-3)·42
Δ = 505
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{505}}{2*-3}=\frac{-1-\sqrt{505}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{505}}{2*-3}=\frac{-1+\sqrt{505}}{-6} $
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