If we apply Rolle's Theorem to the function $f(x)=2 x^{2}-12 x-2$ on the interval $[1,5]$, how many values of $c$ exist such that $f^{\prime}(c)=0$ ?

Step 1 ：First, we need to check if the function satisfies the conditions of Rolle's Theorem. The function \(f(x)=2 x^{2}-12 x-2\) is a polynomial, so it is continuous and differentiable everywhere. We need to check if \(f(1) = f(5)\).

Step 2 ：Calculate \(f(1)\) and \(f(5)\), we find that \(f(1) = f(5) = -12\). So the function satisfies the conditions of Rolle's Theorem on the interval \([1,5]\).

Step 3 ：Next, we need to find the derivative of the function and solve the equation \(f'(c) = 0\) to find the possible values of c. The derivative of the function is \(f'(x) = 4x - 12\).

Step 4 ：Solving the equation \(f'(c) = 0\), we get \(c = 3\). Since \(3\) is in the interval \((1,5)\), there is one value of \(c\) that satisfies the conditions of Rolle's Theorem.

Step 5 ：Final Answer: There is \(\boxed{1}\) value of \(c\) that satisfies the conditions of Rolle's Theorem.