Question

I don't understand how it gets the result
`28b_(2)` and
`b_(2)^(2)` in the fourth equation.

Answer 1

Answer + Step-by-step explanation:

your question is about equation 4 :

`b_(1)=14-b_(2)`

Then

`\left( b_(1)\right)^(2) =\left( 14-b_(2)\right)^(2) \ \text{(this is a special product )}`

Then

`\left( b_(1)\right)^(2) =14^(2)-2* 14 * b_(2)+ (b_2)^2`

Then

`\left( b_(1)\right)^(2) =196-28 b_(2)+ (b_2)^2`