F(2)=2x^2-3

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

(2)=2F^2-3

We move all terms to the left:

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

We get rid of parentheses

-2F^2+3+2=0

We add all the numbers together, and all the variables

-2F^2+5=0

a = -2; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-2)·5
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{-(0)-2\sqrt{10}}{2*-2}=\frac{0-2\sqrt{10}}{-4}
=-\frac{2\sqrt{10}}{-4}
=-\frac{\sqrt{10}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-2}=\frac{0+2\sqrt{10}}{-4}
=\frac{2\sqrt{10}}{-4}
=\frac{\sqrt{10}}{-2}
$