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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-3)·3
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{-(-2)-2\sqrt{10}}{2*-3}=\frac{2-2\sqrt{10}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*-3}=\frac{2+2\sqrt{10}}{-6}
$