f(0)=2,f(2)=4

Solution for f(0)=2,f(2)=4 equation:

f(0)=2.f(2)=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(+2.f^2)+f0=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2.f^2+f=0

We add all the numbers together, and all the variables

-2f^2+f=0

a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·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*-2}=\frac{-2}{-4}
=1/2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2}=\frac{0}{-4}
=0
$