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(F)=F2-3F-5
We move all terms to the left:
(F)-(F2-3F-5)=0
We add all the numbers together, and all the variables
-(+F^2-3F-5)+F=0
We get rid of parentheses
-F^2+3F+F+5=0
We add all the numbers together, and all the variables
-1F^2+4F+5=0
a = -1; b = 4; c = +5;
Δ = b2-4ac
Δ = 42-4·(-1)·5
Δ = 36
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{36}=6$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*-1}=\frac{-10}{-2} =+5 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*-1}=\frac{2}{-2} =-1 $
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