k2+19k+18=0

Solution for k2+19k+18=0 equation:

k2+19k+18=0

We add all the numbers together, and all the variables

k^2+19k+18=0

a = 1; b = 19; c = +18;
Δ = b2-4ac
Δ = 192-4·1·18
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*1}=\frac{-36}{2}
=-18
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*1}=\frac{-2}{2}
=-1
$