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(F)=-2F^2=10F
We move all terms to the left:
(F)-(-2F^2)=0
We get rid of parentheses
2F^2+F=0
a = 2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·2·0
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$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*2}=\frac{-2}{4} =-1/2 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*2}=\frac{0}{4} =0 $
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