0=0.04X^2+19x-500

Solution for 0=0.04X^2+19x-500 equation:

0=0.04X^2+19X-500

We move all terms to the left:

0-(0.04X^2+19X-500)=0

We add all the numbers together, and all the variables

-(0.04X^2+19X-500)=0

We get rid of parentheses

-0.04X^2-19X+500=0

a = -0.04; b = -19; c = +500;
Δ = b2-4ac
Δ = -192-4·(-0.04)·500
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*-0.04}=\frac{-2}{-0.08}
=+25
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*-0.04}=\frac{40}{-0.08}
=-500
$