Question

How do you find the zeros, real and imaginary, of `y = 6x^2 +2x + 6` using the quadratic formula?

Answer 1

There are two imaginary roots at
`x=-1/6+sqrt(35)/6i` and `x=-1/6-sqrt(35)/6i`

Explanation:

The quadratic formula tells us that the general quadratic equation:
`color(white)("XXX")y=ax^2+bx+c`
has zeroes at
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

Given:
`color(white)("XXX")y=6x^2+2x+6`
`a=6`
`b=2`
`c=6`

So the zeroes are at
`color(white)("XXX")x=(-2+-sqrt(2^2-4(6)(6)))/(2(6)`

`color(white)("XXXx")=(-2+-2sqrt(1-36))/(2(6))`

`color(white)("XXXx")=-(1+-sqrt(-35))/6`

`color(white)("XXXx")=-1/6+-sqrt(35)/6*i`