(4-m)x^2 + (2m+4) x + (8m+1)

For the perect square , we know that the discrminant should equal zero.

==> delta = b^2 - 4ac = 0

==> (2m + 4)^2 - 4(4-m)(8m+1) = 0

Let us expand brackets:

==> 4m^2 + 16m + 16 + 32m^2 - 124 m -16 = 0

==> 36m^2 - 108m = 0

==> 36m(m - 3) = 0

==> there is two solutions:

m1= 0

m2= 3

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