(2c-1)(1c-5)=

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

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

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

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

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

Combine like terms: -1c + -10c = -11c
(5 + -11c + 2c2) = 0

Solving
5 + -11c + 2c2 = 0

Solving for variable 'c'.

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

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

Simplifying
5 + -1c = 0

Solving
5 + -1c = 0

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

Add '-5' to each side of the equation.
5 + -5 + -1c = 0 + -5

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

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

Divide each side by '-1'.
c = 5

Simplifying
c = 5
Solution
c = {0.5, 5}