How do you find the roots, real and imaginary, of #y=-4x^2 +x -3# using the quadratic formula?

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Answer 1 · 2018-05-03T04:35:56

#x=(1-isqrt47)/8,# #(1+isqrt47)/8#

The roots are imaginary.

#y=-4x^2+x-3# is a quadratic equation in standard form:

#y=ax^2+x-3#,

where:

#a=-4#, #b=1#, #c=-3#

To find the roots, substitute #0# for #y#.

#0=-4x^2+x-3#

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

Plug in the known values.

#x=(-1+-sqrt(1^2-4*-4*-3))/(2*-4)#

Simplify.

#x=(-1+-sqrt(-47))/(-8)#

#x=(-1+-sqrt(47*-1))/(-8)#

Simplify #sqrt(-1)# to #i#.

#x=(-1+-isqrt47)/(-8)#

#47# is a prime number so it can't be factored.

#x=(1+-isqrt47)/8# #larr# Two negatives make a positive.

Roots

#x=(1-isqrt47)/8,# #(1+isqrt47)/8#

graph{y=-4x^2+x-3 [-15.02, 17, -15.25, 0.77]}