Hi,

N=2x^2+4x+1 , where N, in thousands, have a computer x years after 1990. 

Solve in what year were there 97 thousand households with computers 

 97 = 2x^2 + 4x  1

 2x^2 + 4x - 96 = 0

 x^2 + 2x - 48 = 0

factoring

(x+8)(x-6)= 0

   x = 6 |tossing out negative solution for number of years.

 1990 + 6 = 1996, the year there were 97 thousand households with computers 



 2*6^2 + 4*6 + 1 = 72 + 24 + 1 = 97  |checking our answer 

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