C(x)=-0.00012x^2+0.0100x-0.1=0

Solution for C(x)=-0.00012x^2+0.0100x-0.1=0 equation:

(C)=-0.00012C^2+0.0100C-0.1=0

We move all terms to the left:

(C)-(-0.00012C^2+0.0100C-0.1)=0

We get rid of parentheses

0.00012C^2-0.0100C+C+0.1=0

We add all the numbers together, and all the variables

0.00012C^2+0.99C+0.1=0

a = 0.00012; b = 0.99; c = +0.1;
Δ = b2-4ac
Δ = 0.992-4·0.00012·0.1
Δ = 0.980052
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.99)-\sqrt{0.980052}}{2*0.00012}=\frac{-0.99-\sqrt{0.980052}}{0.00024}
$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.99)+\sqrt{0.980052}}{2*0.00012}=\frac{-0.99+\sqrt{0.980052}}{0.00024}
$