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.