F(2a)=3(2a)^2

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

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

We move all terms to the left:

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

determiningTheFunctionDomain
2F-32F^2=0

We add all the numbers together, and all the variables

-32F^2+2F=0

a = -32; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-32)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-32}=\frac{-4}{-64}
=1/16
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-32}=\frac{0}{-64}
=0
$