Simplify (6t^2-6)/(2t-4)

$\frac{6 t^{2} - 6}{2 t - 4}$

Cancel the common factor of $6 t^{2} - 6$ and $2 t - 4$ .

Factor $2$ out of $6 t^{2}$ .

$\frac{2 (3 t^{2}) - 6}{2 t - 4}$

Factor $2$ out of $- 6$ .

$\frac{2 (3 t^{2}) + 2 \cdot - 3}{2 t - 4}$

Factor $2$ out of $2 (3 t^{2}) + 2 (- 3)$ .

$\frac{2 (3 t^{2} - 3)}{2 t - 4}$

Cancel the common factors.

Factor $2$ out of $2 t$ .

$\frac{2 (3 t^{2} - 3)}{2 (t) - 4}$

Factor $2$ out of $- 4$ .

$\frac{2 (3 t^{2} - 3)}{2 t + 2 \cdot - 2}$

Factor $2$ out of $2 t + 2 \cdot - 2$ .

$\frac{2 (3 t^{2} - 3)}{2 (t - 2)}$

Cancel the common factor.

$\frac{2 (3 t^{2} - 3)}{2 (t - 2)}$

Rewrite the expression.

$\frac{3 t^{2} - 3}{t - 2}$

Simplify the numerator.

Factor $3$ out of $3 t^{2} - 3$ .

Factor $3$ out of $3 t^{2}$ .

$\frac{3 (t^{2}) - 3}{t - 2}$

Factor $3$ out of $- 3$ .

$\frac{3 (t^{2}) + 3 (- 1)}{t - 2}$

Factor $3$ out of $3 (t^{2}) + 3 (- 1)$ .

$\frac{3 (t^{2} - 1)}{t - 2}$

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

$\frac{3 (t^{2} - 1^{2})}{t - 2}$

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

$\frac{3 (t + 1) (t - 1)}{t - 2}$