0,05x^2+2x=0

Solution for 0,05x^2+2x=0 equation:

0.05x^2+2x=0

a = 0.05; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·0.05·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*0.05}=\frac{-4}{0.1}
=-40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*0.05}=\frac{0}{0.1}
=0
$