0=5x^2+19x-4

Solution for 0=5x^2+19x-4 equation:

0=5x^2+19x-4

We move all terms to the left:

0-(5x^2+19x-4)=0

We add all the numbers together, and all the variables

-(5x^2+19x-4)=0

We get rid of parentheses

-5x^2-19x+4=0

a = -5; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·(-5)·4
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*-5}=\frac{-2}{-10}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*-5}=\frac{40}{-10}
=-4
$