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

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

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

a = -4; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·(-4)·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*-4}=\frac{-40}{-8}
=+5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-4}=\frac{2}{-8}
=-1/4
$