S(x)=-0.0003x^2+0.5x+215

Solution for S(x)=-0.0003x^2+0.5x+215 equation:

(S)=-0.0003S^2+0.5S+215

We move all terms to the left:

(S)-(-0.0003S^2+0.5S+215)=0

We get rid of parentheses

0.0003S^2-0.5S+S-215=0

We add all the numbers together, and all the variables

0.0003S^2+0.5S-215=0

a = 0.0003; b = 0.5; c = -215;
Δ = b2-4ac
Δ = 0.52-4·0.0003·(-215)
Δ = 0.508
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.508}}{2*0.0003}=\frac{-0.5-\sqrt{0.508}}{0.0006}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.508}}{2*0.0003}=\frac{-0.5+\sqrt{0.508}}{0.0006}
$