How do you evaluate $(2g - 1) ^2 - 2g + g ^2$ for g = 3?

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Answer 1 · 2017-03-25T17:10:55

For $g=3$ , we have $(2g-1)^2-2g+g^2=28$

Putting $g=3$ in $(2g-1)^2-2g+g^2$ ,

we get $(2xx3-1)^2-2xx3+3^2$

= $(6-1)^2-6+9$

= $5^2+3$

= $25+3$

= $28$

How do you evaluate (2g - 1) ^2 - 2g + g ^2(2g - 1) ^2 - 2g + g ^2 for g = 3?