(k+1)(k+3)=0

Solution for (k+1)(k+3)=0 equation:

(k+1)(k+3)=0

We multiply parentheses ..

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

We get rid of parentheses

k^2+3k+k+3=0

We add all the numbers together, and all the variables

k^2+4k+3=0

a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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