2x+8/x-2+6=0

To solve the equation

\[ \frac{2x + 8}{x - 2} + 6 = 0, \]

we can start by isolating the fraction. First, subtract 6 from both sides:

\[ \frac{2x + 8}{x - 2} = -6. \]

Next, we can eliminate the fraction by multiplying both sides by \(x - 2\), but we need to ensure that \(x \neq 2\) because that would make the denominator zero. Multiplying both sides gives:

\[ 2x + 8 = -6(x - 2). \]

Expanding the right side:

\[ 2x + 8 = -6x + 12. \]

Now, we collect all the \(x\) terms on one side and the constant terms on the other side. Adding \(6x\) to both sides results in:

\[ 2x + 6x + 8 = 12, \]

or

\[ 8x + 8 = 12. \]

Next, we subtract 8 from both sides:

\[ 8x = 4. \]

Dividing both sides by 8, we find:

\[ x = \frac{1}{2}. \]

Thus, the solution to the equation is

\[ \boxed{\frac{1}{2}}. \]

To verify, we can substitute \(x = \frac{1}{2}\) back into the original equation:

\[ \frac{2\left(\frac{1}{2}\right) + 8}{\frac{1}{2} - 2} + 6 = \frac{1 + 8}{\frac{1}{2} - 2} + 6 = \frac{9}{-\frac{3}{2}} + 6 = -6 + 6 = 0. \]

The equation holds true, confirming \(x = \frac{1}{2}\) is the correct solution.