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

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

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

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

Combine like terms: 5c + -2c = 3c
(1 + 3c + -10c2) = 0

Solving
1 + 3c + -10c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(1 + -2c)(1 + 5c) = 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 '(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 = {0.5, -0.2}