(4c+1)(2c+3)=

Solution for (4c+1)(2c+3)= equation:

Simplifying
(4c + 1)(2c + 3) = 0

Reorder the terms:
(1 + 4c)(2c + 3) = 0

Reorder the terms:
(1 + 4c)(3 + 2c) = 0

Multiply (1 + 4c) * (3 + 2c)
(1(3 + 2c) + 4c * (3 + 2c)) = 0
((3 * 1 + 2c * 1) + 4c * (3 + 2c)) = 0
((3 + 2c) + 4c * (3 + 2c)) = 0
(3 + 2c + (3 * 4c + 2c * 4c)) = 0
(3 + 2c + (12c + 8c2)) = 0

Combine like terms: 2c + 12c = 14c
(3 + 14c + 8c2) = 0

Solving
3 + 14c + 8c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(3 + 2c)(1 + 4c) = 0

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

Simplifying
3 + 2c = 0

Solving
3 + 2c = 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 + 2c = 0 + -3

Combine like terms: 3 + -3 = 0
0 + 2c = 0 + -3
2c = 0 + -3

Combine like terms: 0 + -3 = -3
2c = -3

Divide each side by '2'.
c = -1.5

Simplifying
c = -1.5
Subproblem 2
Set the factor '(1 + 4c)' equal to zero and attempt to solve:

Simplifying
1 + 4c = 0

Solving
1 + 4c = 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 + 4c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + 4c = 0 + -1
4c = 0 + -1

Combine like terms: 0 + -1 = -1
4c = -1

Divide each side by '4'.
c = -0.25

Simplifying
c = -0.25
Solution
c = {-1.5, -0.25}