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

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Answer 1 · 2016-05-18T07:09:44

Zeros of #y=-2x^2+x-3# are

#-1/4+isqrt23/4# and #-1/4-isqrt23/4#

If #y=ax^2+bx+c#, quadratic formula states that zeros are given by

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

Hence, in #y=-2x^2+x-3#, zeros are

#(-1+-sqrt(1^2-4*(-2)*(-3)))/(2*2)#

= #(-1+-sqrt(1-24))/(2*2)#

or #(-1+-sqrt(-23))/4=-1/4+-isqrt23/4#