F(x)=2x^2-96=0

Solution for F(x)=2x^2-96=0 equation:

(F)=2F^2-96=0

We move all terms to the left:

(F)-(2F^2-96)=0

We get rid of parentheses

-2F^2+F+96=0

a = -2; b = 1; c = +96;
Δ = b2-4ac
Δ = 12-4·(-2)·96
Δ = 769
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{769}}{2*-2}=\frac{-1-\sqrt{769}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{769}}{2*-2}=\frac{-1+\sqrt{769}}{-4}
$