Explain why the quadratic relation y=2x^2+3x+5 would have no zeros

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.

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation:

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right