k2+19k=0

Solution for k2+19k=0 equation:

k2+19k=0

We add all the numbers together, and all the variables

k^2+19k=0

a = 1; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·1·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*1}=\frac{-38}{2}
=-19
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*1}=\frac{0}{2}
=0
$