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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3F^2-3F-2=0

a = 3; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·3·(-2)
Δ = 33
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{-(-3)-\sqrt{33}}{2*3}=\frac{3-\sqrt{33}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*3}=\frac{3+\sqrt{33}}{6}
$