How do you find the zeros, if any, of #y= 2x^2 +x- 2#using the quadratic formula?

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Answer 1 · 2015-12-24T06:17:49

#x=(-1+-sqrt17)/4#

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

for the quadratic equation

#y=ax^2+bx+c#

Since your equation is

#y=2x^2+x-2#,

#a=2#
#b=1#
#c=-2#

Plug these into the quadratic formula:

#x=(-(1)+-sqrt(1^2-(4xx2xx-2)))/(2xx2)#

#x=(-1+-sqrt(1-(-16)))/4#

#x=(-1+-sqrt17)/4#

These are the two times when the function is equal to #0# (the roots of the function).