F(x)=0.00012x^2+200

Solution for F(x)=0.00012x^2+200 equation:

(F)=0.00012F^2+200

We move all terms to the left:

(F)-(0.00012F^2+200)=0

We get rid of parentheses

-0.00012F^2+F-200=0

a = -0.00012; b = 1; c = -200;
Δ = b2-4ac
Δ = 12-4·(-0.00012)·(-200)
Δ = 0.904
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{0.904}}{2*-0.00012}=\frac{-1-\sqrt{0.904}}{-0.00024}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.904}}{2*-0.00012}=\frac{-1+\sqrt{0.904}}{-0.00024}
$