How do you find the roots, real and imaginary, of #y= -3x^2 -+5x-2# using the quadratic formula?

1 Answer
Alvin L.
Feb 4, 2018

#x_1=6/(-6)=-1#

#x_2=4/(-6)=-2/3#

Explanation:

The quadratic formula states that if you have a quadratic in the form #ax^2+bx+c=0#, the solutions are:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case, #a=-3#, #b=-5# and #c=-2#. We can plug this into the quadratic formula to get:
#x=(-(-5)+-sqrt((-5)^2-4*-3*-2))/(2*-3)#

#x=(5+-sqrt(25-24))/(-6)=(5+-1)/(-6)#

#x_1=6/(-6)=-1#

#x_2=4/(-6)=-2/3#

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