F-2=3x^2+2x-5

Solution for F-2=3x^2+2x-5 equation:

-2=3F^2+2F-5

We move all terms to the left:

-2-(3F^2+2F-5)=0

We get rid of parentheses

-3F^2-2F+5-2=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}
$