F(x)=3x^2+2x+1,x=1

Solution for F(x)=3x^2+2x+1,x=1 equation:

(F)=3F^2+2F+1.F=1

We move all terms to the left:

(F)-(3F^2+2F+1.F)=0

We get rid of parentheses

-3F^2+F-2F-1.F=0

We add all the numbers together, and all the variables

-3F^2-2F=0

a = -3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-3}=\frac{0}{-6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-3}=\frac{4}{-6}
=-2/3
$