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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-3)·3
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{5}}{2*-3}=\frac{3-3\sqrt{5}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{5}}{2*-3}=\frac{3+3\sqrt{5}}{-6}
$