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

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

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

We move all terms to the left:

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

We get rid of parentheses

-3F^2+F+5=0

a = -3; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-3)·5
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{61}}{2*-3}=\frac{-1-\sqrt{61}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{61}}{2*-3}=\frac{-1+\sqrt{61}}{-6}
$