k^2+9k=-20

Solution for k^2+9k=-20 equation:

k^2+9k=-20

We move all terms to the left:

k^2+9k-(-20)=0

We add all the numbers together, and all the variables

k^2+9k+20=0

a = 1; b = 9; c = +20;
Δ = b2-4ac
Δ = 92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*1}=\frac{-10}{2}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*1}=\frac{-8}{2}
=-4
$