F(x)=3x^2+12x=1

Solution for F(x)=3x^2+12x=1 equation:

(F)=3F^2+12F=1

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2-11F=0

a = -3; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-3)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*-3}=\frac{0}{-6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-3}=\frac{22}{-6}
=-3+2/3
$