Solve for $\mathrm{x}$.
\[
\begin{array}{c}
2 x^{2}-28 x+98=0 \\
x=[?]
\end{array}
\]
Remember the quadratic formula: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Step 1 ：Given the quadratic equation \(2x^{2}-28x+98=0\).

Step 2 ：Identify the coefficients as \(a = 2\), \(b = -28\), and \(c = 98\).

Step 3 ：Use the quadratic formula to solve for \(x\), which is \(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\).

Step 4 ：Substitute the values of \(a\), \(b\), and \(c\) into the formula.

Step 5 ：Calculate the discriminant \(D = b^{2}-4ac\). In this case, \(D = 0\).

Step 6 ：Since the discriminant is zero, the equation has a repeated root.

Step 7 ：Calculate the roots \(x1 = \frac{-b + \sqrt{D}}{2a} = 7.0\) and \(x2 = \frac{-b - \sqrt{D}}{2a} = 7.0\).

Step 8 ：Final Answer: The solution to the equation is \(\boxed{7}\).