x^2+0.4x-0.04=0

Solution for x^2+0.4x-0.04=0 equation:

x^2+0.4x-0.04=0

a = 1; b = 0.4; c = -0.04;
Δ = b2-4ac
Δ = 0.42-4·1·(-0.04)
Δ = 0.32
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{-(0.4)-\sqrt{0.32}}{2*1}=\frac{-0.4-\sqrt{0.32}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.32}}{2*1}=\frac{-0.4+\sqrt{0.32}}{2}
$