f(1)=6,f(2)=12,f(3)=24,f(4)=48

Solution for f(1)=6,f(2)=12,f(3)=24,f(4)=48 equation:

f(1)=6.f(2)=12.f(3)=24.f(4)=48

We move all terms to the left:

f(1)-(6.f(2))=0

We add all the numbers together, and all the variables

-(+6.f^2)+f1=0

We add all the numbers together, and all the variables

-(+6.f^2)+f=0

We get rid of parentheses

-6.f^2+f=0

We add all the numbers together, and all the variables

-6f^2+f=0

a = -6; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-6)·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*-6}=\frac{-2}{-12}
=1/6
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-6}=\frac{0}{-12}
=0
$