To find the range of values for which \(y < 0\) for the equation \(y = 2x^2 + 9x - 35\), we need to solve for the values of \(x\) that make \(y\) less than zero.

First, set \(y\) to zero and solve for \(x\):

\(2x^2 + 9x - 35 = 0\)

Now, factor the quadratic equation:

\((2x - 5)(x + 7) = 0\)

Now, set each factor equal to zero and solve for \(x\):

1. \(2x - 5 = 0\)
  \(2x = 5\)
  \(x = \frac{5}{2}\)

2. \(x + 7 = 0\)
  \(x = -7\)

So, we have two critical points: \(x = \frac{5}{2}\) and \(x = -7\).

Now, let's examine the intervals around these critical points to determine the range of values for which \(y < 0\).

1. For \(x < -7\): Plug in a value less than -7 into the original equation (e.g., \(x = -8\)):
  \(y = 2(-8)^2 + 9(-8) - 35 = 64 - 72 - 35 = -43\)
  So, for \(x < -7\), \(y\) is less than zero.

2. For \(-7 < x < \frac{5}{2}\): Plug in a value between -7 and \(\frac{5}{2}\) into the original equation (e.g., \(x = 0\)):
  \(y = 2(0)^2 + 9(0) - 35 = 0 - 0 - 35 = -35\)
  So, for \(-7 < x < \frac{5}{2}\), \(y\) is less than zero.

3. For \(x > \frac{5}{2}\): Plug in a value greater than \(\frac{5}{2}\) into the original equation (e.g., \(x = 3\)):
  \(y = 2(3)^2 + 9(3) - 35 = 18 + 27 - 35 = 10\)
  So, for \(x > \frac{5}{2}\), \(y\) is greater than zero.

Now, we can see that the range of values for which \(y < 0\) is \(-7 < x < \frac{5}{2}\).