c(c+2)-c(c-6)=10c-128

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Solution for c(c+2)-c(c-6)=10c-128 equation: 


Simplifying
c(c + 2) + -1c(c + -6) = 10c + -128

Reorder the terms:
c(2 + c) + -1c(c + -6) = 10c + -128
(2 * c + c * c) + -1c(c + -6) = 10c + -128
(2c + c2) + -1c(c + -6) = 10c + -128

Reorder the terms:
2c + c2 + -1c(-6 + c) = 10c + -128
2c + c2 + (-6 * -1c + c * -1c) = 10c + -128
2c + c2 + (6c + -1c2) = 10c + -128

Reorder the terms:
2c + 6c + c2 + -1c2 = 10c + -128

Combine like terms: 2c + 6c = 8c
8c + c2 + -1c2 = 10c + -128

Combine like terms: c2 + -1c2 = 0
8c + 0 = 10c + -128
8c = 10c + -128

Reorder the terms:
8c = -128 + 10c

Solving
8c = -128 + 10c

Solving for variable 'c'.

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

Add '-10c' to each side of the equation.
8c + -10c = -128 + 10c + -10c

Combine like terms: 8c + -10c = -2c
-2c = -128 + 10c + -10c

Combine like terms: 10c + -10c = 0
-2c = -128 + 0
-2c = -128

Divide each side by '-2'.
c = 64

Simplifying
c = 64

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