Complete the square to rewrite the quadratic:
2 x² + 3 x + 5 = 2 (x² + 3/2 x) + 5
... = 2 (x² + 3/2 x + 9/16 - 9/16) + 5
... = 2 (x² + 3/2 x + (3/4)²) + 5 - 9/8
... = 2 (x + 3/4)² + 31/8
Any real number squared becomes non-negative, so the quadratic expression has a minimum value of 31/8, which is greater than 0, and so there are no (real) x for which y = 0.
