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

1 Answer
The Legend
Feb 21, 2018

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

Related questions
See all questions in Quadratic Formula
Impact of this question
2109 views around the world
You can reuse this answer
Creative Commons License