Find all values of $k$ for which the quadratic equation has two real solutions.
\[
2 x^{2}+7 x+k=0
\]
Write your answer as an equality or inequality in terms of $k$.

Step 1 ：Given the quadratic equation \(2x^{2}+7x+k=0\), we need to find all values of \(k\) for which the equation has two real solutions.

Step 2 ：A quadratic equation has two real solutions if and only if the discriminant of the equation is greater than 0. The discriminant of a quadratic equation in the form \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\).

Step 3 ：In this case, \(a = 2\), \(b = 7\), and \(c = k\). So, we need to find all values of \(k\) such that \(7^2 - 4*2*k > 0\).

Step 4 ：Solving the inequality \(49 - 8k > 0\) gives us \(k < \frac{49}{8}\).

Step 5 ：\(\boxed{k < \frac{49}{8}}\) is the final answer. The quadratic equation has two real solutions for all values of \(k\) that are less than \(\frac{49}{8}\).