f(0+2)=f(0+1)+4f(0)

Solution for f(0+2)=f(0+1)+4f(0) equation:

f(0+2)=f(0+1)+4f(0)

We move all terms to the left:

f(0+2)-(f(0+1)+4f(0))=0

We add all the numbers together, and all the variables

f2-(f1+4f0)=0

We add all the numbers together, and all the variables

f^2-(f1+4f0)=0

We get rid of parentheses

f^2-f1-4f0=0

We add all the numbers together, and all the variables

f^2-5f=0

a = 1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*1}=\frac{0}{2}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*1}=\frac{10}{2}
=5
$