Wrong extra hint on end of solving quads by fact.

Fixes Open Khan#33388 Khan#33076 Khan#33077 Closed Khan#31645 Khan#29435 Khan#30472 Khan#30400 Khan#27614 Solving quadratics by factoring (& 2) added and extra hint for the one-root problem. Both used appendContents but recently multiple problem had one more hint than one-root problem added. Kept the new text, but merged the new hint with old.
karlanderson · Nov 10, 2012 · fe5ed18 · fe5ed18
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diff --git a/exercises/solving_quadratics_by_factoring.html b/exercises/solving_quadratics_by_factoring.html
@@ -56,13 +56,13 @@
                     So <code>(x <var>A &lt; 0 ? "+" : ""</var> \color{<var>PINK</var>}{<var>-A</var>})
                     (x <var>B &lt; 0 ? "+" : ""</var> \color{<var>PINK</var>}{<var>-B</var>}) = 0</code>.
                 </p>
-                <p id="hint3">
-                    Since <code>(x <var>A &lt; 0 ? "+" : ""</var> <var>-A</var>)
-                    (x <var>B &lt; 0 ? "+" : ""</var> <var>-B</var>) = 0</code>,
-                    we know that one or both quantities must equal zero for the equation to be true.
-                </p>
-                <p id="hint4"><code>x + <var>-A</var> = 0</code> or <code>x + <var>-B</var> = 0</code></p>
-                <p id="hint5"><b>Thus, <code>x = <var>A</var></code> and <code>x = <var>B</var></code> are the solutions.</b></p>
+                <div id="hint3">
+                    <p>Since <code>(x <var>A &lt; 0 ? "+" : ""</var> <var>-A</var>)
+                        (x <var>B &lt; 0 ? "+" : ""</var> <var>-B</var>) = 0</code>,
+                        we know that one or both quantities must equal zero for the equation to be true.</p>
+                    <p><code>x + <var>-A</var> = 0</code> or <code>x + <var>-B</var> = 0</code></p>
+                </div>
+                <p id="hint4"><b>Thus, <code>x = <var>A</var></code> and <code>x = <var>B</var></code> are the solutions.</b></p>
             </div>
         </div>
 

diff --git a/exercises/solving_quadratics_by_factoring_2.html b/exercises/solving_quadratics_by_factoring_2.html
@@ -56,13 +56,13 @@
                     So <code>(x <var>A &lt; 0 ? "+" : ""</var> \color{<var>PINK</var>}{<var>-A</var>})
                     (x <var>B &lt; 0 ? "+" : ""</var> \color{<var>PINK</var>}{<var>-B</var>}) = 0</code>.
                 </p>
-                <p id="hint3">
-                    Since <code>(x <var>A &lt; 0 ? "+" : ""</var> <var>-A</var>)
-                    (x <var>B &lt; 0 ? "+" : ""</var> <var>-B</var>) = 0</code>,
-                    we know that one or both quantities must equal zero for the equation to be true.
-                </p>
-                <p id="hint4"><code>x + <var>-A</var> = 0</code> or <code>x + <var>-B</var> = 0</code></p>
-                <p id="hint5"><b>Thus, <code>x = <var>A</var></code> and <code>x = <var>B</var></code> are the solutions.</b></p>
+                <div id="hint3">
+                    <p>Since <code>(x <var>A &lt; 0 ? "+" : ""</var> <var>-A</var>)
+                        (x <var>B &lt; 0 ? "+" : ""</var> <var>-B</var>) = 0</code>,
+                        we know that one or both quantities must equal zero for the equation to be true.</p>
+                    <p><code>x + <var>-A</var> = 0</code> or <code>x + <var>-B</var> = 0</code></p>
+                </div>
+                <p id="hint4"><b>Thus, <code>x = <var>A</var></code> and <code>x = <var>B</var></code> are the solutions.</b></p>
             </div>
         </div>