3(3c+1)(3c-1)=21c-5

Solution for 3(3c+1)(3c-1)=21c-5 equation:

Simplifying
3(3c + 1)(3c + -1) = 21c + -5

Reorder the terms:
3(1 + 3c)(3c + -1) = 21c + -5

Reorder the terms:
3(1 + 3c)(-1 + 3c) = 21c + -5

Multiply (1 + 3c) * (-1 + 3c)
3(1(-1 + 3c) + 3c * (-1 + 3c)) = 21c + -5
3((-1 * 1 + 3c * 1) + 3c * (-1 + 3c)) = 21c + -5
3((-1 + 3c) + 3c * (-1 + 3c)) = 21c + -5
3(-1 + 3c + (-1 * 3c + 3c * 3c)) = 21c + -5
3(-1 + 3c + (-3c + 9c2)) = 21c + -5

Combine like terms: 3c + -3c = 0
3(-1 + 0 + 9c2) = 21c + -5
3(-1 + 9c2) = 21c + -5
(-1 * 3 + 9c2 * 3) = 21c + -5
(-3 + 27c2) = 21c + -5

Reorder the terms:
-3 + 27c2 = -5 + 21c

Solving
-3 + 27c2 = -5 + 21c

Solving for variable 'c'.

Reorder the terms:
-3 + 5 + -21c + 27c2 = -5 + 21c + 5 + -21c

Combine like terms: -3 + 5 = 2
2 + -21c + 27c2 = -5 + 21c + 5 + -21c

Reorder the terms:
2 + -21c + 27c2 = -5 + 5 + 21c + -21c

Combine like terms: -5 + 5 = 0
2 + -21c + 27c2 = 0 + 21c + -21c
2 + -21c + 27c2 = 21c + -21c

Combine like terms: 21c + -21c = 0
2 + -21c + 27c2 = 0

Factor a trinomial.
(1 + -9c)(2 + -3c) = 0

Subproblem 1
Set the factor '(1 + -9c)' equal to zero and attempt to solve:

Simplifying
1 + -9c = 0

Solving
1 + -9c = 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 + -9c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -9c = 0 + -1
-9c = 0 + -1

Combine like terms: 0 + -1 = -1
-9c = -1

Divide each side by '-9'.
c = 0.1111111111

Simplifying
c = 0.1111111111
Subproblem 2
Set the factor '(2 + -3c)' equal to zero and attempt to solve:

Simplifying
2 + -3c = 0

Solving
2 + -3c = 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 + -3c = 0 + -2

Combine like terms: 2 + -2 = 0
0 + -3c = 0 + -2
-3c = 0 + -2

Combine like terms: 0 + -2 = -2
-3c = -2

Divide each side by '-3'.
c = 0.6666666667

Simplifying
c = 0.6666666667
Solution
c = {0.1111111111, 0.6666666667}