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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2F^2-3F+7=0

a = -2; b = -3; c = +7;
Δ = b2-4ac
Δ = -32-4·(-2)·7
Δ = 65
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{65}}{2*-2}=\frac{3-\sqrt{65}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*-2}=\frac{3+\sqrt{65}}{-4}
$