Solve for n (2-n)(6+2/3n)(n-2)=0

$(2 - n) (6 + \frac{2}{3} n) (n - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 - n = 0$

$6 + \frac{2}{3} n = 0$

$n - 2 = 0$

Set $2 - n$ equal to $0$ and solve for $n$ .

Set $2 - n$ equal to $0$ .

$2 - n = 0$

Solve $2 - n = 0$ for $n$ .

Subtract $2$ from both sides of the equation.

$- n = - 2$

Divide each term in $- n = - 2$ by $- 1$ and simplify.

Divide each term in $- n = - 2$ by $- 1$ .

$\frac{- n}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{n}{1} = \frac{- 2}{- 1}$

Divide $n$ by $1$ .

$n = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$n = 2$

Set $6 + \frac{2}{3} n$ equal to $0$ and solve for $n$ .

Set $6 + \frac{2}{3} n$ equal to $0$ .

$6 + \frac{2}{3} n = 0$

Solve $6 + \frac{2}{3} n = 0$ for $n$ .

Combine $\frac{2}{3}$ and $n$ .

$6 + \frac{2 n}{3} = 0$

Subtract $6$ from both sides of the equation.

$\frac{2 n}{3} = - 6$

Multiply both sides of the equation by $\frac{3}{2}$ .

$\frac{3}{2} \cdot \frac{2 n}{3} = \frac{3}{2} \cdot - 6$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{3}{2} \cdot \frac{2 n}{3}$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{2} \cdot \frac{2 n}{3} = \frac{3}{2} \cdot - 6$

Rewrite the expression.

$\frac{1}{2} (2 n) = \frac{3}{2} \cdot - 6$

Cancel the common factor of $2$ .

Factor $2$ out of $2 n$ .

$\frac{1}{2} (2 (n)) = \frac{3}{2} \cdot - 6$

Cancel the common factor.

$\frac{1}{2} (2 n) = \frac{3}{2} \cdot - 6$

Rewrite the expression.

$n = \frac{3}{2} \cdot - 6$

Simplify the right side.

Simplify $\frac{3}{2} \cdot - 6$ .

Cancel the common factor of $2$ .

Factor $2$ out of $- 6$ .

$n = \frac{3}{2} \cdot (2 (- 3))$

Cancel the common factor.

$n = \frac{3}{2} \cdot (2 \cdot - 3)$

Rewrite the expression.

$n = 3 \cdot - 3$

Multiply $3$ by $- 3$ .

$n = - 9$

Set $n - 2$ equal to $0$ and solve for $n$ .

Set $n - 2$ equal to $0$ .

$n - 2 = 0$

Add $2$ to both sides of the equation.

$n = 2$

The final solution is all the values that make $(2 - n) (6 + \frac{2}{3} n) (n - 2) = 0$ true.

$n = 2, - 9$

Solve for n (2-n)(6+2/3n)(n-2)=0

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