F(2)=x2-4x+5

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Solution for F(2)=x2-4x+5 equation:



(2)=F2-4F+5
We move all terms to the left:
(2)-(F2-4F+5)=0
We add all the numbers together, and all the variables
-(+F^2-4F+5)+2=0
We get rid of parentheses
-F^2+4F-5+2=0
We add all the numbers together, and all the variables
-1F^2+4F-3=0
a = -1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-1}=\frac{-6}{-2} =+3 $
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-1}=\frac{-2}{-2} =1 $

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