(k-9)(k+5)=0

Solution for (k-9)(k+5)=0 equation:

(k-9)(k+5)=0

We multiply parentheses ..

(+k^2+5k-9k-45)=0

We get rid of parentheses

k^2+5k-9k-45=0

We add all the numbers together, and all the variables

k^2-4k-45=0

a = 1; b = -4; c = -45;
Δ = b2-4ac
Δ = -42-4·1·(-45)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*1}=\frac{-10}{2}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*1}=\frac{18}{2}
=9
$