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Find and explain the error in the​ student’s work/answer below.

​Solve: ​ 2c² + c​ – 6​ = 0
         ​ 2c² + 4c​ – 3c ​ – 6​ = 0
      ​ 2c(c +​ 2) –​ 3(c +​ 2) = 0
                ​ (c+2)(2c –​ 3) = 0
The student states that the answers are c​ + 2 and 2c​ – 3.

The error in the student's work is that the student treats the terms 4c and -3c as like terms and combines them to get c. This is incorrect, as these terms cannot be combined because they do not have the same variable part. The correct way to solve this equation is to use the quadratic formula, factoring, or completing the square. The correct solutions are c = -2 and c = 3/2. Therefore, the student's answers of c + 2 and 2c - 3 are both incorrect.