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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4c^2+8c+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:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*4}=\frac{-12}{8}
=-1+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*4}=\frac{-4}{8}
=-1/2
$