F(3y)=(3y)^2

Solution for F(3y)=(3y)^2 equation:

(3F)=(3F)^2

We move all terms to the left:

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

determiningTheFunctionDomain
3F-3F^2=0

We add all the numbers together, and all the variables

-3F^2+3F=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$