0.04x^2+0.01x-5=0

Solution for 0.04x^2+0.01x-5=0 equation:

0.04x^2+0.01x-5=0

a = 0.04; b = 0.01; c = -5;
Δ = b2-4ac
Δ = 0.012-4·0.04·(-5)
Δ = 0.8001
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{0.8001}}{2*0.04}=\frac{-0.01-\sqrt{0.8001}}{0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.01)+\sqrt{0.8001}}{2*0.04}=\frac{-0.01+\sqrt{0.8001}}{0.08}
$