F(x)=x^2-4x-96

Solution for F(x)=x^2-4x-96 equation:

(F)=F^2-4F-96

We move all terms to the left:

(F)-(F^2-4F-96)=0

We get rid of parentheses

-F^2+F+4F+96=0

We add all the numbers together, and all the variables

-1F^2+5F+96=0

a = -1; b = 5; c = +96;
Δ = b2-4ac
Δ = 52-4·(-1)·96
Δ = 409
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{-(5)-\sqrt{409}}{2*-1}=\frac{-5-\sqrt{409}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{409}}{2*-1}=\frac{-5+\sqrt{409}}{-2}
$