How do you find the zeros, real and imaginary, of #y=-2x^2-9x+5# using the quadratic formula?

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Answer 1 · 2018-04-22T11:04:29

Zeros are real and #x=0.5, x= -5#

# y= -2 x^2-9 x +5#

Comparing with standard quadratic equation #ax^2+bx+c=0#

# a=-2 ,b=-9 ,c=5#. Discriminant # D= b^2-4 a c# or

#D=81+40 =121# , discriminant is positive, we get two real

solutions. Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (9+-sqrt 121)/(-4) = (9 +- 11)/-4 :. x= 20/-4= -5 # or

#x=(-2)/-4 = 0.5 :. # Zeros are real and #x=0.5, x= -5# [Ans]