x^2+0.2x-0.04x=0

Solution for x^2+0.2x-0.04x=0 equation:

x^2+0.2x-0.04x=0

We add all the numbers together, and all the variables

x^2+0.16x=0

a = 1; b = 0.16; c = 0;
Δ = b2-4ac
Δ = 0.162-4·1·0
Δ = 0.0256
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{-(0.16)-\sqrt{0.0256}}{2*1}=\frac{-0.16-\sqrt{0.0256}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.16)+\sqrt{0.0256}}{2*1}=\frac{-0.16+\sqrt{0.0256}}{2}
$