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

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Answer 1 · 2015-11-23T20:28:11

This equation has 2 real zeros:

#x_1=(-1+sqrt(101))/10#

#x_2=(-1-sqrt(101))/10#

To find zeros of quadratic equation you have to calculate #Delta# first:

#Delta=b^2-4ac=(-1)^2-4*(-5)*5=1+100=101#

#Delta >0# so the formula has 2 real zeros.

#x_1=(-b-sqrt(Delta))/(2a)=(1-sqrt(101))/(-10)=(-1+sqrt(101))/10#

#x_2=(-b+sqrt(Delta))/(2a)=(1+sqrt(101))/(-10)=(-1-sqrt(101))/10#

If #Delta<0# then there would be 2 imaginary zeros (2 complex conjugate numbers)

If #Delta=0# there would be a single real zero