diff --git a/other/clrs/07/04/03.markdown b/other/clrs/07/04/03.markdown
index 21435baa..0cc8ece4 100644
--- a/other/clrs/07/04/03.markdown
+++ b/other/clrs/07/04/03.markdown
@@ -1,13 +1,14 @@
 > Show that the expression $q^2 + (n - q - 1)^2$ achieves a maximum over $q =
 > 0, 1, \ldots, n-1$ when $q = 0$ and $q = n - 1$.
 
+
 $$ \begin{align}
    f(q)   &= q^2 + (n - q - 1)^2 \\\\
    f'(q)  &= 2q - 2(n - q - 1) = 4q - 2n + 2 \\\\
    f''(q) &= 4 \\\\
    \end{align} $$
 
-$f'(q) = 0$ when $q = \frac{1}{2}n - \frac{1}{4}$. $f'(q)$ is also continious.
+$f'(q) = 0$ when $q = \frac{1}{2}n - \frac{1}{2}$. $f'(q)$ is also continuous.
 $\forall q : f''(q) > 0$, which means that $f'(q)$ is negative left of $f'(q) =
 0$ and positive right of it, which means that this is a local minima. In this
 case, $f(q)$ is decreasing in the beginning of the interval and increasing in