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

1 Answer
Feb 21, 2018

Answer:

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#