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

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

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

Reorder the terms:
(1 + 2c)(c + 3) = 0

Reorder the terms:
(1 + 2c)(3 + c) = 0

Multiply (1 + 2c) * (3 + c)
(1(3 + c) + 2c * (3 + c)) = 0
((3 * 1 + c * 1) + 2c * (3 + c)) = 0
((3 + 1c) + 2c * (3 + c)) = 0
(3 + 1c + (3 * 2c + c * 2c)) = 0
(3 + 1c + (6c + 2c2)) = 0

Combine like terms: 1c + 6c = 7c
(3 + 7c + 2c2) = 0

Solving
3 + 7c + 2c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(3 + c)(1 + 2c) = 0

Subproblem 1
Set the factor '(3 + c)' equal to zero and attempt to solve:

Simplifying
3 + c = 0

Solving
3 + c = 0

Move all terms containing c to the left, all other terms to the right.

Add '-3' to each side of the equation.
3 + -3 + c = 0 + -3

Combine like terms: 3 + -3 = 0
0 + c = 0 + -3
c = 0 + -3

Combine like terms: 0 + -3 = -3
c = -3

Simplifying
c = -3
Subproblem 2
Set the factor '(1 + 2c)' equal to zero and attempt to solve:

Simplifying
1 + 2c = 0

Solving
1 + 2c = 0

Move all terms containing c to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + 2c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + 2c = 0 + -1
2c = 0 + -1

Combine like terms: 0 + -1 = -1
2c = -1

Divide each side by '2'.
c = -0.5

Simplifying
c = -0.5
Solution
c = {-3, -0.5}