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f(0)=-f^2+3f+1
We move all terms to the left:
f(0)-(-f^2+3f+1)=0
We add all the numbers together, and all the variables
-(-f^2+3f+1)+f=0
We get rid of parentheses
f^2-3f+f-1=0
We add all the numbers together, and all the variables
f^2-2f-1=0
a = 1; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·1·(-1)
Δ = 8
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*1}=\frac{2-2\sqrt{2}}{2} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*1}=\frac{2+2\sqrt{2}}{2} $
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