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(1)=F2-5F+6
We move all terms to the left:
(1)-(F2-5F+6)=0
We add all the numbers together, and all the variables
-(+F^2-5F+6)+1=0
We get rid of parentheses
-F^2+5F-6+1=0
We add all the numbers together, and all the variables
-1F^2+5F-5=0
a = -1; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·(-1)·(-5)
Δ = 5
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{5}}{2*-1}=\frac{-5-\sqrt{5}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*-1}=\frac{-5+\sqrt{5}}{-2} $
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