(k+3)(k+7)=0

Solution for (k+3)(k+7)=0 equation:

(k+3)(k+7)=0

We multiply parentheses ..

(+k^2+7k+3k+21)=0

We get rid of parentheses

k^2+7k+3k+21=0

We add all the numbers together, and all the variables

k^2+10k+21=0

a = 1; b = 10; c = +21;
Δ = b2-4ac
Δ = 102-4·1·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*1}=\frac{-14}{2}
=-7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*1}=\frac{-6}{2}
=-3
$