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(F)=-3F^2-30F-80
We move all terms to the left:
(F)-(-3F^2-30F-80)=0
We get rid of parentheses
3F^2+30F+F+80=0
We add all the numbers together, and all the variables
3F^2+31F+80=0
a = 3; b = 31; c = +80;
Δ = b2-4ac
Δ = 312-4·3·80
Δ = 1
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*3}=\frac{-32}{6} =-5+1/3 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*3}=\frac{-30}{6} =-5 $
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