(k+9)(k+2)=-12

Solution for (k+9)(k+2)=-12 equation:

(k+9)(k+2)=-12

We move all terms to the left:

(k+9)(k+2)-(-12)=0

We add all the numbers together, and all the variables

(k+9)(k+2)+12=0

We multiply parentheses ..

(+k^2+2k+9k+18)+12=0

We get rid of parentheses

k^2+2k+9k+18+12=0

We add all the numbers together, and all the variables

k^2+11k+30=0

a = 1; b = 11; c = +30;
Δ = b2-4ac
Δ = 112-4·1·30
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{-(11)-1}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*1}=\frac{-10}{2}
=-5
$