(2k+1)(2k+3)=0

Solution for (2k+1)(2k+3)=0 equation:

(2k+1)(2k+3)=0

We multiply parentheses ..

(+4k^2+6k+2k+3)=0

We get rid of parentheses

4k^2+6k+2k+3=0

We add all the numbers together, and all the variables

4k^2+8k+3=0

a = 4; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·4·3
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{-(8)-4}{2*4}=\frac{-12}{8}
=-1+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*4}=\frac{-4}{8}
=-1/2
$