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(F)=5F^2+2F
We move all terms to the left:
(F)-(5F^2+2F)=0
We get rid of parentheses
-5F^2+F-2F=0
We add all the numbers together, and all the variables
-5F^2-1F=0
a = -5; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-5)·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$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-5}=\frac{0}{-10} =0 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-5}=\frac{2}{-10} =-1/5 $
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