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

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

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

a = -0.00012; b = 0.0100; c = -0.1;
Δ = b2-4ac
Δ = 0.01002-4·(-0.00012)·(-0.1)
Δ = 5.2E-5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.0100)-\sqrt{5.2E-5}}{2*-0.00012}=\frac{-0.01-\sqrt{5.2E-5}}{-0.00024}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.0100)+\sqrt{5.2E-5}}{2*-0.00012}=\frac{-0.01+\sqrt{5.2E-5}}{-0.00024}
$