jmc

Moving all the terms to the LHS, we have the equation $x^2-14x+y^2-48y=0$. Completing the square on the quadratic in $x$, we add $(14/2{)}^2=49$ to both sides. Completing the square on the quadratic in $y$, we add $(48/2{)}^2=576$ to both sides. We have the equation $$(x^2-14x+49)+(y^2-48y+576)=625\Rightarrow (x-7{)}^2+(y-24{)}^2=625$$Rearranging, we have $(x-7{)}^2=625-(y-24{)}^2$. Taking the square root and solving for $x$, we get $x=\pm \sqrt{625-(y-24{)}^2}+7$. Since $\sqrt{625-(y-24{)}^2}$ is always nonnegative, the minimum value of $x$ is achieved when we use a negative sign in front of the square root. Now, we want the largest possible value of the square root. In other words, we want to maximize $625-(y-24{)}^2$. Since $(y-24{)}^2$ is always nonnegative, $625-(y-24{)}^2$ is maximized when $(y-24{)}^2=0$ or when $y=24$. At this point, $625-(y-24{)}^2=625$ and $x=-\sqrt{625}+7=-18$. Thus, the minimum $x$ value is $\boxed{-18}$.

--OR--

Similar to the solution above, we can complete the square to get the equation $(x-7{)}^2+(y-24{)}^2=625$. This equation describes a circle with center at $(7,24)$ and radius $\sqrt{625}=25$. The minimum value of $x$ is achieved at the point on the left side of the circle, which is located at $(7-25,24)=(-18,24)$. Thus, the minimum value of $x$ is $\boxed{-18}$.