Solve for $\mathrm{x}$. \[ \begin{array}{c} 2 x^{2}-4 x-48=0 \\ x=[?], \square \end{array} \] Enter the smallest solution first. Remember the quadratic formula: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Step 1 ：Given the quadratic equation \(2x^{2} - 4x - 48 = 0\), we can solve for x using the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).

Step 2 ：Here, a = 2, b = -4, and c = -48.

Step 3 ：Substitute these values into the quadratic formula to find the solutions for x.

Step 4 ：Calculate the discriminant D = \(b^{2} - 4ac = (-4)^{2} - 4*2*(-48) = 400\).

Step 5 ：Since the discriminant D is positive, there are two real and distinct solutions for x.

Step 6 ：Calculate the first solution: \(x_{1} = \frac{-(-4) - \sqrt{400}}{2*2} = -4.0\).

Step 7 ：Calculate the second solution: \(x_{2} = \frac{-(-4) + \sqrt{400}}{2*2} = 6.0\).

Step 8 ：The solutions to the equation are \(x = \boxed{-4.0}\) and \(x = \boxed{6.0}\).