Help me

$\frac{3x+12}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}$

First, you can take a 3 out of the numerator on the left.

$\frac{3(x+4)}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}$

Combine like factors on the top right.

$\frac{3(x+4)}{x+4} = \frac{12x^2+8x}{2x}$

The x+4's on the right can cancel.

$3=\frac{12x^2+8x}{2x}$

The left can be divided by 2x.

$3=6x+4$

Subtract 4 from both sides.

$-1=6x$

Divide both sides by 6.

$x=-\frac{1}{6}$

x=-1/6