f(1)=-1,f(2)=3,f(4)=-1

Solution for f(1)=-1,f(2)=3,f(4)=-1 equation:

f(1)=-1.f(2)=3.f(4)=-1

We move all terms to the left:

f(1)-(-1.f(2))=0

We add all the numbers together, and all the variables

-(-1f^2)+f1=0

We add all the numbers together, and all the variables

-(-1f^2)+f=0

We get rid of parentheses

1f^2+f=0

We add all the numbers together, and all the variables

f^2+f=0

a = 1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*1}=\frac{-2}{2}
=-1
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*1}=\frac{0}{2}
=0
$