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

Solution for F(0)=3x^2+4x-2 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -3; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·(-3)·2
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-3}=\frac{4-2\sqrt{10}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-3}=\frac{4+2\sqrt{10}}{-6}
$