Question

How do you find the zeros, real and imaginary, of `y=5x^2+7x+12` using the quadratic formula?

Answer 1

`x=-7/10+-sqrt(191)/10i`

Explanation:

let's solve the equation
`5x^2+7x+12=0`
using the quadratic formula, so
`x=(-7+-sqrt(49-240))/10`
under root we have a negative number, so the solutions are not real
`x=-7/10-sqrt(191)/10i`
and
`x=-7/10+sqrt(191)/10i`