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

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

-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{-40}{-10}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-5}=\frac{2}{-10}
=-1/5
$