2x^2+0.2x-0.01=0

Solution for 2x^2+0.2x-0.01=0 equation:

2x^2+0.2x-0.01=0

a = 2; b = 0.2; c = -0.01;
Δ = b2-4ac
Δ = 0.22-4·2·(-0.01)
Δ = 0.12
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{-(0.2)-\sqrt{0.12}}{2*2}=\frac{-0.2-\sqrt{0.12}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{0.12}}{2*2}=\frac{-0.2+\sqrt{0.12}}{4}
$