F(x)=-2x^2+5

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

(F)=-2F^2+5

We move all terms to the left:

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

We get rid of parentheses

2F^2+F-5=0

a = 2; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·2·(-5)
Δ = 41
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{-(1)-\sqrt{41}}{2*2}=\frac{-1-\sqrt{41}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*2}=\frac{-1+\sqrt{41}}{4}
$