F(x)=3x^2+12x+8

Solution for F(x)=3x^2+12x+8 equation:

(F)=3F^2+12F+8

We move all terms to the left:

(F)-(3F^2+12F+8)=0

We get rid of parentheses

-3F^2+F-12F-8=0

We add all the numbers together, and all the variables

-3F^2-11F-8=0

a = -3; b = -11; c = -8;
Δ = b2-4ac
Δ = -112-4·(-3)·(-8)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*-3}=\frac{6}{-6}
=-1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*-3}=\frac{16}{-6}
=-2+2/3
$