(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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