How do you find the zeros, real and imaginary, of #y= -7x^2-3x+2 # using the quadratic formula?

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Answer 1 · 2018-01-26T05:49:09

About #-0.79# and #0.36#.

This quadratic is in standard form: #ax^2+bx+c#

So for #y=-7x^2-3x+2#,
#a=-7#,
#b=-3#, and
#c=2#.

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

and plug in our #a#, #b#, and #c# values and simplify:

#x=(3+-sqrt((-3)^2-4(-7)(2)))/(2(-7))#

#=(3+-sqrt(9+56))/(-14)#

#=(3+-sqrt(65))/(-14)#

which splits into the two solutions

#x=(3+sqrt(65))/(-14) ~~ -0.79#
and
#x=(3-sqrt(65))/(-14) ~~ 0.36#