-0.01x^2-0.4x+12=0

Solution for -0.01x^2-0.4x+12=0 equation:

-0.01x^2-0.4x+12=0

a = -0.01; b = -0.4; c = +12;
Δ = b2-4ac
Δ = -0.42-4·(-0.01)·12
Δ = 0.64
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{-(-0.4)-\sqrt{0.64}}{2*-0.01}=\frac{0.4-\sqrt{0.64}}{-0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.4)+\sqrt{0.64}}{2*-0.01}=\frac{0.4+\sqrt{0.64}}{-0.02}
$