Solve by Factoring n^2-1+6n=6n

$n^{2} - 1 + 6 n = 6 n$

Subtract $6 n$ from both sides of the equation.

$n^{2} - 1 + 6 n - 6 n = 0$

Combine the opposite terms in $n^{2} - 1 + 6 n - 6 n$ .

Subtract $6 n$ from $6 n$ .

$n^{2} - 1 + 0 = 0$

Add $n^{2} - 1$ and $0$ .

$n^{2} - 1 = 0$

Rewrite $1$ as $1^{2}$ .

$n^{2} - 1^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = n$ and $b = 1$ .

$(n + 1) (n - 1) = 0$

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

$n + 1 = 0$

$n - 1 = 0$

Set $n + 1$ equal to $0$ and solve for $n$ .

Set $n + 1$ equal to $0$ .

$n + 1 = 0$

Subtract $1$ from both sides of the equation.

$n = - 1$

Set $n - 1$ equal to $0$ and solve for $n$ .

Set $n - 1$ equal to $0$ .

$n - 1 = 0$

Add $1$ to both sides of the equation.

$n = 1$

The final solution is all the values that make $(n + 1) (n - 1) = 0$ true.

$n = - 1, 1$