F(x)=3x^2+9x+4

Solution for F(x)=3x^2+9x+4 equation:

(F)=3F^2+9F+4

We move all terms to the left:

(F)-(3F^2+9F+4)=0

We get rid of parentheses

-3F^2+F-9F-4=0

We add all the numbers together, and all the variables

-3F^2-8F-4=0

a = -3; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·(-3)·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*-3}=\frac{4}{-6}
=-2/3
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*-3}=\frac{12}{-6}
=-2
$