Let $A$ be an $n\times n$ complex matrix with $n\geq 3$. It is shown that at least $n-2$ of the eigenvalues of $A$ lie in the disk \begin{equation*}\left\vert z-\frac{\func{tr}A}{n}\right\vert \leq \sqrt{\frac{n-1}{n}\left(\sqrt{\left( \left\Vert A\right\Vert _{2}^{2}-\frac{\left\vert \func{tr} A\right\vert ^{2}}{n}\right) ^{2}-\frac{\left\Vert A^{\ast }A-AA^{\ast}\right\Vert _{2}^{2}}{2}}-\frac{\limfunc{spd}\nolimits^{2}(A)}{2}\right) },\end{equation*} where $\left\Vert A\right\Vert _{2},$ $\func{tr}A$, and $\limfunc{spd}(A)$ denote the Frobenius norm, the trace, and the spread of $A$, respectively. In particular, if $A=\left[ a_{ij}\right] $ is normal, then at least $n-2$ of the eigenvalues of $A$ lie in the disk {\small \begin{eqnarray*} & & \left\vert z-\frac{\func{tr}A}{n}\right\vert \\ & & \leq \sqrt{\frac{n-1}{n}\left( \frac{\left\Vert A\right\Vert _{2}^{2}}{2}-\frac{\left\vert \func{tr}A\right\vert ^{2}}{n}-\frac{3}{2}\max_{i,j=1,\dots,n} \left( \sum_{\substack{ k=1 \\ k\neq i}}^{n}\left\vert a_{ki}\right\vert ^{2}+\sum_{\substack{ k=1 \\ k\neq j}}^{n}\left\vert a_{kj}\right\vert ^{2}+\frac{\left\vert a_{ii}-a_{jj}\right\vert ^{2}}{2}\right) \right) }. \end{eqnarray*}} Moreover, the constant $\frac{3}{2}$ can be replaced by $4$ if the matrix $A$ is Hermitian.