(4c+1)(4c+1)=

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

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

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

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

Multiply (1 + 4c) * (1 + 4c)
(1(1 + 4c) + 4c * (1 + 4c)) = 0
((1 * 1 + 4c * 1) + 4c * (1 + 4c)) = 0
((1 + 4c) + 4c * (1 + 4c)) = 0
(1 + 4c + (1 * 4c + 4c * 4c)) = 0
(1 + 4c + (4c + 16c2)) = 0

Combine like terms: 4c + 4c = 8c
(1 + 8c + 16c2) = 0

Solving
1 + 8c + 16c2 = 0

Solving for variable 'c'.

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

Subproblem 1
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
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 = {-0.25, -0.25}