x^2+0.01x-0.01=0

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

x^2+0.01x-0.01=0

a = 1; b = 0.01; c = -0.01;
Δ = b2-4ac
Δ = 0.012-4·1·(-0.01)
Δ = 0.0401
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{-(0.01)-\sqrt{0.0401}}{2*1}=\frac{-0.01-\sqrt{0.0401}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.01)+\sqrt{0.0401}}{2*1}=\frac{-0.01+\sqrt{0.0401}}{2}
$