F(x)=-0.01x^2+0.6x+5.3

Solution for F(x)=-0.01x^2+0.6x+5.3 equation:

(F)=-0.01F^2+0.6F+5.3

We move all terms to the left:

(F)-(-0.01F^2+0.6F+5.3)=0

We get rid of parentheses

0.01F^2-0.6F+F-5.3=0

We add all the numbers together, and all the variables

0.01F^2+0.4F-5.3=0

a = 0.01; b = 0.4; c = -5.3;
Δ = b2-4ac
Δ = 0.42-4·0.01·(-5.3)
Δ = 0.372
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{-(0.4)-\sqrt{0.372}}{2*0.01}=\frac{-0.4-\sqrt{0.372}}{0.02}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.372}}{2*0.01}=\frac{-0.4+\sqrt{0.372}}{0.02}
$