k2+9k+18=0

Solution for k2+9k+18=0 equation:

k2+9k+18=0

We add all the numbers together, and all the variables

k^2+9k+18=0

a = 1; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·1·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*1}=\frac{-6}{2}
=-3
$