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.

option a is correct.

only one solution to the system of equations

step-by-step explanation:

if the graph of the equations do not intersects(i.,e if they are parallel) then, there are no solutions that are true for both the equations.

if the graph of the equations intersects, then there is one solution and that is true for both the equations.

since, in the given figure:

the two lines intersect each other

⇒there is only one solution which satisfy both the lines of an equations.

therefore, there is only one solution to the system of equation

the answer will be 80.