6-(2c-1)=(3+2c)(3-2c)

Solution for 6-(2c-1)=(3+2c)(3-2c) equation:

Simplifying
6 + -1(2c + -1) = (3 + 2c)(3 + -2c)

Reorder the terms:
6 + -1(-1 + 2c) = (3 + 2c)(3 + -2c)
6 + (-1 * -1 + 2c * -1) = (3 + 2c)(3 + -2c)
6 + (1 + -2c) = (3 + 2c)(3 + -2c)

Combine like terms: 6 + 1 = 7
7 + -2c = (3 + 2c)(3 + -2c)

Multiply (3 + 2c) * (3 + -2c)
7 + -2c = (3(3 + -2c) + 2c * (3 + -2c))
7 + -2c = ((3 * 3 + -2c * 3) + 2c * (3 + -2c))
7 + -2c = ((9 + -6c) + 2c * (3 + -2c))
7 + -2c = (9 + -6c + (3 * 2c + -2c * 2c))
7 + -2c = (9 + -6c + (6c + -4c2))

Combine like terms: -6c + 6c = 0
7 + -2c = (9 + 0 + -4c2)
7 + -2c = (9 + -4c2)

Solving
7 + -2c = 9 + -4c2

Solving for variable 'c'.

Reorder the terms:
7 + -9 + -2c + 4c2 = 9 + -4c2 + -9 + 4c2

Combine like terms: 7 + -9 = -2
-2 + -2c + 4c2 = 9 + -4c2 + -9 + 4c2

Reorder the terms:
-2 + -2c + 4c2 = 9 + -9 + -4c2 + 4c2

Combine like terms: 9 + -9 = 0
-2 + -2c + 4c2 = 0 + -4c2 + 4c2
-2 + -2c + 4c2 = -4c2 + 4c2

Combine like terms: -4c2 + 4c2 = 0
-2 + -2c + 4c2 = 0

Factor out the Greatest Common Factor (GCF), '2'.
2(-1 + -1c + 2c2) = 0

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

Ignore the factor 2.
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 + -1c)' equal to zero and attempt to solve:

Simplifying
1 + -1c = 0

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

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

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

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

Simplifying
c = 1
Solution
c = {-0.5, 1}