To find the set of \(x\) values for which \(y > 9\), start with the inequality:

\(2x^2 - 3x > 9\)

Subtract 9 from both sides:

\(2x^2 - 3x - 9 > 0\)

Factor the quadratic expression:

\((2x + 3)(x - 3) > 0\)

Find the roots by setting each factor to zero:

\(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)

\(x - 3 = 0 \Rightarrow x = 3\)

Test intervals around the roots to determine where the inequality holds:

• For \(x < -\frac{3}{2}\), choose \(x = -2\): \((2(-2) + 3)((-2) - 3) = (-1)(-5) = 5 > 0\)
• For \(-\frac{3}{2} < x < 3\), choose \(x = 0\): \((2(0) + 3)(0 - 3) = (3)(-3) = -9 < 0\)
• For \(x > 3\), choose \(x = 4\): \((2(4) + 3)(4 - 3) = (11)(1) = 11 > 0\)

The solution is \(x > 3\) or \(x < -\frac{3}{2}\).