Pt hoành độ giao điểm
\(-\dfrac{1}{2}x^2=2x-6\\↔x^2=-4x+12\\↔x^2+4x-12=0\\↔x^2+6x-2x-12=0\\↔x(x+6)-2(x+6)=0\\↔(x-2)(x+6)=0\\↔\left[\begin{array}{1}x-2=0\\x+6=0\end{array}\right.\\↔\left[\begin{array}{1}x=2\\x=-6\end{array}\right.\)
mà \(x_A>0\\→x_A=2\\→y_A=-2\\→A(2;-2)\)
\( (d) \) đi qua \(A(2;-2)\)
\(→3.2+2m-4=-2\\↔2m+2=-2\\↔2m=-4\\↔m=-2\)
Vậy \(m=-2\)