3(3c+5)+1=2(c-20)

Solution for 3(3c+5)+1=2(c-20) equation:

Simplifying
3(3c + 5) + 1 = 2(c + -20)

Reorder the terms:
3(5 + 3c) + 1 = 2(c + -20)
(5 * 3 + 3c * 3) + 1 = 2(c + -20)
(15 + 9c) + 1 = 2(c + -20)

Reorder the terms:
15 + 1 + 9c = 2(c + -20)

Combine like terms: 15 + 1 = 16
16 + 9c = 2(c + -20)

Reorder the terms:
16 + 9c = 2(-20 + c)
16 + 9c = (-20 * 2 + c * 2)
16 + 9c = (-40 + 2c)

Solving
16 + 9c = -40 + 2c

Solving for variable 'c'.

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

Add '-2c' to each side of the equation.
16 + 9c + -2c = -40 + 2c + -2c

Combine like terms: 9c + -2c = 7c
16 + 7c = -40 + 2c + -2c

Combine like terms: 2c + -2c = 0
16 + 7c = -40 + 0
16 + 7c = -40

Add '-16' to each side of the equation.
16 + -16 + 7c = -40 + -16

Combine like terms: 16 + -16 = 0
0 + 7c = -40 + -16
7c = -40 + -16

Combine like terms: -40 + -16 = -56
7c = -56

Divide each side by '7'.
c = -8

Simplifying
c = -8