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.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

the answer is c just got it on ed

step-by-step explanation: