19k^2+10k=0

Solution for 19k^2+10k=0 equation:

19k^2+10k=0

a = 19; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·19·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*19}=\frac{-20}{38}
=-10/19
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*19}=\frac{0}{38}
=0
$