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.