Question

How do you find the zeros, real and imaginary, of `y=x^2+3x+7` using the quadratic formula?

Answer 1

Plug the equation into the formula and solve for x. These will be your zeroes.

Explanation:

`y=ax^2+bx+c`
`x=(-b+-sqrt(b^2-4ac))/(2a)`
plug in the values from the equation
`x=(-3+-sqrt(3^2-4(1)(7)))/(2(1))`
then solve for x...
`x=-3/2+-sqrt(-19)/2`

`x=(-3+-isqrt19)/2`
This gives 2 imaginary zeroes, but does not give any real zeroes, and if you graph the equation you will see that the equation does not have any.