(k+9)(k+4)=0

Solution for (k+9)(k+4)=0 equation:

(k+9)(k+4)=0

We multiply parentheses ..

(+k^2+4k+9k+36)=0

We get rid of parentheses

k^2+4k+9k+36=0

We add all the numbers together, and all the variables

k^2+13k+36=0

a = 1; b = 13; c = +36;
Δ = b2-4ac
Δ = 132-4·1·36
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*1}=\frac{-18}{2}
=-9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*1}=\frac{-8}{2}
=-4
$