19w^2+33w=0

Solution for 19w^2+33w=0 equation:

19w^2+33w=0

a = 19; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·19·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*19}=\frac{-66}{38}
=-1+14/19
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*19}=\frac{0}{38}
=0
$