F(3a)=3a^2-3(3a)+5

Solution for F(3a)=3a^2-3(3a)+5 equation:

(3F)=3F^2-3(3F)+5

We move all terms to the left:

(3F)-(3F^2-3(3F)+5)=0

We get rid of parentheses

-3F^2+3F+33F-5=0

We add all the numbers together, and all the variables

-3F^2+36F-5=0

a = -3; b = 36; c = -5;
Δ = b2-4ac
Δ = 362-4·(-3)·(-5)
Δ = 1236
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{309}}{2*-3}=\frac{-36-2\sqrt{309}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{309}}{2*-3}=\frac{-36+2\sqrt{309}}{-6}
$