Use the fundamental theorem of algebra to determine the number of roots for 2x^2+4x+7

Altho' I'm not using the fund. thm. of alg. specifically to determine the # of roots of 2x^2 + 4x + 7, polynomials of the nth degree all have n roots.

Completing the square:  2x^2 + 4x                                 + 7
                                       2(x^2 + 2x + 1   -   1)                +7
                                        2(x+1)^2           - 2                    +7
                                        2(x+1)^2 + 5

To solve for the roots, set the above = to 0 and solve for x:

2(x+1)^2 = -5   =>   (x+1)^2 = -5/2
 
 x+1 = plus or minus sqrt (-5/2)       => x+1 = plus or minus i*sqrt(5/2)

... and so on.  As expected, this 2nd order poly has 2 roots.  The roots in this case happen to be complex.                

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