-0.04x^2+0.4x+1.9=0

Solution for -0.04x^2+0.4x+1.9=0 equation:

-0.04x^2+0.4x+1.9=0

a = -0.04; b = 0.4; c = +1.9;
Δ = b2-4ac
Δ = 0.42-4·(-0.04)·1.9
Δ = 0.464
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{0.464}}{2*-0.04}=\frac{-0.4-\sqrt{0.464}}{-0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.464}}{2*-0.04}=\frac{-0.4+\sqrt{0.464}}{-0.08}
$