0.02x2-0.06x+0.01=0

Solution for 0.02x2-0.06x+0.01=0 equation:

0.02x^2-0.06x+0.01=0

a = 0.02; b = -0.06; c = +0.01;
Δ = b2-4ac
Δ = -0.062-4·0.02·0.01
Δ = 0.0028
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{-(-0.06)-\sqrt{0.0028}}{2*0.02}=\frac{0.06-\sqrt{0.0028}}{0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.06)+\sqrt{0.0028}}{2*0.02}=\frac{0.06+\sqrt{0.0028}}{0.04}
$