c(3c+2)-3(c^2+4c-17)=1

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Solution for c(3c+2)-3(c^2+4c-17)=1 equation: 


Simplifying
c(3c + 2) + -3(c2 + 4c + -17) = 1

Reorder the terms:
c(2 + 3c) + -3(c2 + 4c + -17) = 1
(2 * c + 3c * c) + -3(c2 + 4c + -17) = 1
(2c + 3c2) + -3(c2 + 4c + -17) = 1

Reorder the terms:
2c + 3c2 + -3(-17 + 4c + c2) = 1
2c + 3c2 + (-17 * -3 + 4c * -3 + c2 * -3) = 1
2c + 3c2 + (51 + -12c + -3c2) = 1

Reorder the terms:
51 + 2c + -12c + 3c2 + -3c2 = 1

Combine like terms: 2c + -12c = -10c
51 + -10c + 3c2 + -3c2 = 1

Combine like terms: 3c2 + -3c2 = 0
51 + -10c + 0 = 1
51 + -10c = 1

Solving
51 + -10c = 1

Solving for variable 'c'.

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

Add '-51' to each side of the equation.
51 + -51 + -10c = 1 + -51

Combine like terms: 51 + -51 = 0
0 + -10c = 1 + -51
-10c = 1 + -51

Combine like terms: 1 + -51 = -50
-10c = -50

Divide each side by '-10'.
c = 5

Simplifying
c = 5

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