Question

I don't understand how it gets the result 8b_{2}" align="absmiddle" class="latex-formula"> 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

$\left( b_{1}\right)^{2} =14^{2}-2\times 14 \times b_{2}+ (b_2)^2$

Then

$\left( b_{1}\right)^{2} =196-28 b_{2}+ (b_2)^2$