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(F)=F^2-12F-29
We move all terms to the left:
(F)-(F^2-12F-29)=0
We get rid of parentheses
-F^2+F+12F+29=0
We add all the numbers together, and all the variables
-1F^2+13F+29=0
a = -1; b = 13; c = +29;
Δ = b2-4ac
Δ = 132-4·(-1)·29
Δ = 285
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{-(13)-\sqrt{285}}{2*-1}=\frac{-13-\sqrt{285}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{285}}{2*-1}=\frac{-13+\sqrt{285}}{-2} $
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