Algebra 1

$m=-\frac{3}{2}, 4$
Given: $3(m+4)=2m(m-1)$ $3m+12=2m^2-2m$ $2m^2-2m-3m-12=0$ $2m^2-5m-12=0$ $2m^2+8m-3m-12=0$ $(2m+3)(m-4)=0$ Solutions are: $2m+3=0 \implies m=-\frac{3}{2}$, and $m-4=0 \implies m=4$