(5c+2)(5c+1)=

Solution for (5c+2)(5c+1)= equation:

Simplifying
(5c + 2)(5c + 1) = 0

Reorder the terms:
(2 + 5c)(5c + 1) = 0

Reorder the terms:
(2 + 5c)(1 + 5c) = 0

Multiply (2 + 5c) * (1 + 5c)
(2(1 + 5c) + 5c * (1 + 5c)) = 0
((1 * 2 + 5c * 2) + 5c * (1 + 5c)) = 0
((2 + 10c) + 5c * (1 + 5c)) = 0
(2 + 10c + (1 * 5c + 5c * 5c)) = 0
(2 + 10c + (5c + 25c2)) = 0

Combine like terms: 10c + 5c = 15c
(2 + 15c + 25c2) = 0

Solving
2 + 15c + 25c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(2 + 5c)(1 + 5c) = 0

Subproblem 1
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
Subproblem 2
Set the factor '(1 + 5c)' equal to zero and attempt to solve:

Simplifying
1 + 5c = 0

Solving
1 + 5c = 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 + 5c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + 5c = 0 + -1
5c = 0 + -1

Combine like terms: 0 + -1 = -1
5c = -1

Divide each side by '5'.
c = -0.2

Simplifying
c = -0.2
Solution
c = {-0.4, -0.2}