I don't understand how it gets the result [tex]28b_{2}[/tex] and [tex]b_{2}^{2}[/tex] in the fourth equation.


Answer + Step-by-step explanation:
your question is about equation 4 :
[tex]b_{1}=14-b_{2}[/tex]
Then
[tex]\left( b_{1}\right)^{2} =\left( 14-b_{2}\right)^{2} \ \text{(this is a special product )}[/tex]
Then
[tex]\left( b_{1}\right)^{2} =14^{2}-2\times 14 \times b_{2}+ (b_2)^2[/tex]
Then
[tex]\left( b_{1}\right)^{2} =196-28 b_{2}+ (b_2)^2[/tex]