(х+2)(2х-8)-14=0

To solve the equation \((x+2)(2x-8) - 14 = 0\), we first expand the left side of the equation:

\((x+2)(2x-8) - 14 = 2x^2 - 8x + 4x - 16 - 14 = 2x^2 - 4x - 30\)

Now, the equation becomes:

\(2x^2 - 4x - 30 = 0\)

To solve this quadratic equation, we can factor out a common factor of 2:

\(2(x^2 - 2x - 15) = 0\)

Now, we factor the quadratic expression inside the parentheses:

\(x^2 - 2x - 15 = (x-5)(x+3) = 0\)

Setting each factor to zero, we find two possible solutions:

\(x-5 = 0 \implies x = 5\)

\(x+3 = 0 \implies x = -3\)

Therefore, the solutions to the equation \((x+2)(2x-8) - 14 = 0\) are \(x = 5\) and \(x = -3\).