F(3x)=(3x)^2-3x

Solution for F(3x)=(3x)^2-3x equation:

(3F)=(3F)^2-3F

We move all terms to the left:

(3F)-((3F)^2-3F)=0

We get rid of parentheses

-3F^2+3F+3F=0

We add all the numbers together, and all the variables

-3F^2+6F=0

a = -3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-3)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-3}=\frac{-12}{-6}
=+2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-3}=\frac{0}{-6}
=0
$