F(n)=2n^2-10

Solution for F(n)=2n^2-10 equation:

(F)=2F^2-10

We move all terms to the left:

(F)-(2F^2-10)=0

We get rid of parentheses

-2F^2+F+10=0

a = -2; b = 1; c = +10;
Δ = b2-4ac
Δ = 12-4·(-2)·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*-2}=\frac{8}{-4}
=-2
$