Complete the square to rewrite the quadratic:
2 <em>x</em>² + 3 <em>x</em> + 5 = 2 (<em>x</em>² + 3/2 <em>x</em>) + 5
... = 2 (<em>x</em>² + 3/2 <em>x</em> + 9/16 - 9/16) + 5
... = 2 (<em>x</em>² + 3/2 <em>x</em> + (3/4)²) + 5 - 9/8
... = 2 (<em>x</em> + 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) <em>x</em> for which <em>y</em> = 0.
