F(20)=50-0.03x^2

Solution for F(20)=50-0.03x^2 equation:

(20)=50-0.03F^2

We move all terms to the left:

(20)-(50-0.03F^2)=0

We get rid of parentheses

0.03F^2-50+20=0

We add all the numbers together, and all the variables

0.03F^2-30=0

a = 0.03; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·0.03·(-30)
Δ = 3.6
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{-(0)-\sqrt{3.6}}{2*0.03}=\frac{0-\sqrt{3.6}}{0.06}
=-\frac{\sqrt{}}{0.06}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{3.6}}{2*0.03}=\frac{0+\sqrt{3.6}}{0.06}
=\frac{\sqrt{}}{0.06}
$