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(F)=-2F^2+6F
We move all terms to the left:
(F)-(-2F^2+6F)=0
We get rid of parentheses
2F^2-6F+F=0
We add all the numbers together, and all the variables
2F^2-5F=0
a = 2; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·2·0
Δ = 25
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{25}=5$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*2}=\frac{0}{4} =0 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*2}=\frac{10}{4} =2+1/2 $
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