How do you find the zeros, real and imaginary, of #y=-4x^2-2x+33# using the quadratic formula?

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Answer 1 · 2017-04-16T00:33:53

#x~~-3.133# and #~~2.633#

The quadratic formula is bases on #ax^2+bx+c#:
#(-b+-sqrt(b^2-(4*a*c)))/(2*a)#

So, our equation of #y=-4x^2-2x+33# tells us #a=-4#, #b=-2#, #c=33#

Thus: #(-(-2)+-sqrt((-2)^2-(4*-4*33)))/(2*-4)#

#(2+-sqrt(4-(4*-4*33)))/(-8)#

#(2+-sqrt(4--528))/-8#

#(2+-sqrt532)/(-8)#

#(2+-2sqrt133)/(-8)#

#(2(1+-sqrt133))/(-8)#

#(1+-sqrt133)/-4#

That's the exact form, but the estimated form is #x~~-3.133# and #~~2.633#