(5c+2)(c+2)=0

Solution for (5c+2)(c+2)=0 equation:

Simplifying
(5c + 2)(c + 2) = 0

Reorder the terms:
(2 + 5c)(c + 2) = 0

Reorder the terms:
(2 + 5c)(2 + c) = 0

Multiply (2 + 5c) * (2 + c)
(2(2 + c) + 5c * (2 + c)) = 0
((2 * 2 + c * 2) + 5c * (2 + c)) = 0
((4 + 2c) + 5c * (2 + c)) = 0
(4 + 2c + (2 * 5c + c * 5c)) = 0
(4 + 2c + (10c + 5c2)) = 0

Combine like terms: 2c + 10c = 12c
(4 + 12c + 5c2) = 0

Solving
4 + 12c + 5c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(2 + c)(2 + 5c) = 0

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

Simplifying
2 + c = 0

Solving
2 + c = 0

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

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

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

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

Simplifying
c = -2
Subproblem 2
Set the factor '(2 + 5c)' equal to zero and attempt to solve:

Simplifying
2 + 5c = 0

Solving
2 + 5c = 0

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

Add '-2' to each side of the equation.
2 + -2 + 5c = 0 + -2

Combine like terms: 2 + -2 = 0
0 + 5c = 0 + -2
5c = 0 + -2

Combine like terms: 0 + -2 = -2
5c = -2

Divide each side by '5'.
c = -0.4

Simplifying
c = -0.4
Solution
c = {-2, -0.4}