.0001x^2+.04x-1.6=0

Solution for .0001x^2+.04x-1.6=0 equation:

.0001x^2+.04x-1.6=0

a = .0001; b = .04; c = -1.6;
Δ = b2-4ac
Δ = .042-4·.0001·(-1.6)
Δ = 0.00224
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{-(.04)-\sqrt{0.00224}}{2*.0001}=\frac{-0.04-\sqrt{0.00224}}{0.0002}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.04)+\sqrt{0.00224}}{2*.0001}=\frac{-0.04+\sqrt{0.00224}}{0.0002}
$