F(x)=3x^2+18x+9

Solution for F(x)=3x^2+18x+9 equation:

(F)=3F^2+18F+9

We move all terms to the left:

(F)-(3F^2+18F+9)=0

We get rid of parentheses

-3F^2+F-18F-9=0

We add all the numbers together, and all the variables

-3F^2-17F-9=0

a = -3; b = -17; c = -9;
Δ = b2-4ac
Δ = -172-4·(-3)·(-9)
Δ = 181
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{-(-17)-\sqrt{181}}{2*-3}=\frac{17-\sqrt{181}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{181}}{2*-3}=\frac{17+\sqrt{181}}{-6}
$