The quadratic has one root with multiplicity 2 if the discriminant is 0, which is
[tex](2m+4)^2-4(4-m)(8m+1)[/tex]
(That is, for a quadratic [tex]ax^2+bx+c[/tex], the discriminant is [tex]b^2-4ac[/tex].)
Set the discriminant equal to 0 and solve for [tex]m[/tex]:
[tex](2m+4)^2-4(4-m)(8m+1)=0[/tex]
[tex]4m^2+16m+16+32m^2-124m-16=0[/tex]
[tex]36m^2-108=0[/tex]
[tex]36m(m-3)=0[/tex]
[tex]\implies m=0\text{ or }m=3[/tex]
