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

1 Answer
Nov 27, 2015

Answer:

# x = +1 1/2 color(white)(.........)x=-1#

#{EEx : AA x in RR}color(white)(....)# Calculating y is just a matter of substituting for #x#. However it is obvious that #y=0#

Explanation:

Given:#color(white)(..)y=-2x^2+x+3#

Standard form equation: #color(white)(..)y=ax^2+bx+c#

#color(white)(.....) x= (-b+-sqrt(b^2-4ac))/(2a)#

In this case:

#a=(-2)#
#b=1#
#c=3#

So#color(white)(....) x= (-1+-sqrt(1^2-4(-2)(3)))/(2(-2))#

#color(white)(....) x= (-1+-sqrt(1+24))/(-4)#

#x=(-1+-5)/(-4)#

# x = +1 1/2 color(white)(.........)x=-1#
Tont B