F(x+1)=3x^2+4x-3

Solution for F(x+1)=3x^2+4x-3 equation:

(F+1)=3F^2+4F-3

We move all terms to the left:

(F+1)-(3F^2+4F-3)=0

We get rid of parentheses

-3F^2+F-4F+1+3=0

We add all the numbers together, and all the variables

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

a = -3; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-3)·4
Δ = 57
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{57}}{2*-3}=\frac{3-\sqrt{57}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-3}=\frac{3+\sqrt{57}}{-6}
$