F(-2)=x^2+4x-8

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Solution for F(-2)=x^2+4x-8 equation:



(-2)=F^2+4F-8
We move all terms to the left:
(-2)-(F^2+4F-8)=0
We add all the numbers together, and all the variables
-(F^2+4F-8)-2=0
We get rid of parentheses
-F^2-4F+8-2=0
We add all the numbers together, and all the variables
-1F^2-4F+6=0
a = -1; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·(-1)·6
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-1}=\frac{4-2\sqrt{10}}{-2} $
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-1}=\frac{4+2\sqrt{10}}{-2} $

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