Question

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

Answer 1

There are two real zeroes:
`color(white)("XXX")x=-4/7+sqrt(22)/7color(white)("XX")` and `color(white)("XX")x=-4/7-sqrt(22)/7`

Explanation:

The quadratic formula for an expression of the form
`color(white)("XXX")ax^2+bx+c`
tells us that the zeroes occur at
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

For the given expression: `-7x^2-2x+3`
`color(white)("XXX")a=-7`
`color(white)("XXX")b=-2`
`color(white)("XXX")c=+3`

So the zeroes are at
`color(white)("XXX")x=(-(-2)+-sqrt((-2)^2+4(-7)(3)))/(2(-7))`
`color(white)("XXX")=(4+-sqrt(4+84))/(-14)`
`color(white)("XXX")=-(4+-2sqrt(22))/14`
`color(white)("XXX")=-2/7+-sqrt(22)/7`

Note that since the argument of the square root is positive,
the zeroes are Real.