RE #11689 Fixing rst problem

Code/Mantid/docs/source/algorithms/EstimatePeakErrors-v1.rst

@@ -13,7 +13,7 @@ This algorithm takes a function after it has been optimized by the Fit algorithm
 
 Usage
 -----
-(*At the moment the algorithm works properly if run from C++ only.*)
+(*At the moment the algorithm works properly if run from C++ only.*)::
 
     import numpy as np
 
@@ -34,5 +34,5 @@ Usage
     # Calculate peak parameter error estimates for the two Lorentzians.
     params = EstimatePeakErrors(fun)
 
- 
+
 .. categories::

Code/Mantid/docs/source/algorithms/LRScalingFactors-v1.rst

@@ -25,17 +25,16 @@ no attenuators.
 
 The normalization run for a data set taken in a given slit setting
 configuration can then be expressed in terms of the standard 0-attenuator
-data set with:
-   D_i = F_i D_0
-
-  Here's an example of runs and how they are related to F.
-
-        run: 55889, att: 0, s1: 0.26, s2: 0.26
-        run: 55890, att: 1, s1: 0.26, s2: 0.26
-        run: 55891, att: 1, s1: 0.33, s2: 0.26 --> F = 55891 / 55890
-        run: 55892, att: 1, s1: 0.45, s2: 0.26 --> F = 55892 / 55890
-        run: 55895, att: 1, s1: 0.81, s2: 0.26
-        run: 55896, att: 2, s1: 0.81, s2: 0.26
-        run: 55897, att: 2, s1: 1.05, s2: 0.35 --> F = 55897 / 55896 * 55895 / 55890
+data set with: :math:`D_i = F_i D_0` 
+
+Here's an example of runs and how they are related to F.
+
+-  run: 55889, att: 0, s1: 0.26, s2: 0.26
+-  run: 55890, att: 1, s1: 0.26, s2: 0.26
+-  run: 55891, att: 1, s1: 0.33, s2: 0.26 --> F = 55891 / 55890
+-  run: 55892, att: 1, s1: 0.45, s2: 0.26 --> F = 55892 / 55890
+-  run: 55895, att: 1, s1: 0.81, s2: 0.26
+-  run: 55896, att: 2, s1: 0.81, s2: 0.26
+-  run: 55897, att: 2, s1: 1.05, s2: 0.35 --> F = 55897 / 55896 * 55895 / 55890
 
 .. categories::