Question

How do you find the zeros, real and imaginary, of `y= -7x^2-5x+3` using the quadratic formula?

Answer 1

The zeroes (or roots, as they are also called) of `y=-7x^2-5x+3` are `x=(5+-sqrt(109))/-14`

Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is `x=(-b+-sqrt(b^2-4ac))/(2a)`. `a`,`b`, and `c` come from `ax^2+bx+c`, the standard form of a quadratic equation.

The equation is already in standard form (thank goodness). `a=-7`, `b=-5`, and `c=3`.
Plugging these into the quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(-5)+-sqrt((-5)^2-4(-7)(3)))/(2(-7))`
`x=(5+-sqrt(25-(-28)(3)))/-14`
`x=(5+-sqrt(25-(-84)))/-14`
`x=(5+-sqrt(25+84))/-14`
`x=(5+-sqrt(109))/-14`
109 is a prime number, so `sqrt(109)` is in simplest form. Therefore our answer is `x=(5+-sqrt(109))/-14`