F(3)=2x^2-5

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

(3)=2F^2-5

We move all terms to the left:

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

We get rid of parentheses

-2F^2+5+3=0

We add all the numbers together, and all the variables

-2F^2+8=0

a = -2; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-2)·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-2}=\frac{-8}{-4}
=+2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-2}=\frac{8}{-4}
=-2
$