0.01x^2+2x+1=0

Solution for 0.01x^2+2x+1=0 equation:

0.01x^2+2x+1=0

a = 0.01; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·0.01·1
Δ = 3.96
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{-(2)-\sqrt{3.96}}{2*0.01}=\frac{-2-\sqrt{3.96}}{0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{3.96}}{2*0.01}=\frac{-2+\sqrt{3.96}}{0.02}
$