F(x)=x^2-3x-108

Solution for F(x)=x^2-3x-108 equation:

(F)=F^2-3F-108

We move all terms to the left:

(F)-(F^2-3F-108)=0

We get rid of parentheses

-F^2+F+3F+108=0

We add all the numbers together, and all the variables

-1F^2+4F+108=0

a = -1; b = 4; c = +108;
Δ = b2-4ac
Δ = 42-4·(-1)·108
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{7}}{2*-1}=\frac{-4-8\sqrt{7}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{7}}{2*-1}=\frac{-4+8\sqrt{7}}{-2}
$