thay x=3 vào pt ta được 

\(12-2\left(2-3\right)^2=4\left(3-m\right)-\left(3-3\right)\left(2x+5\right)\)

\(12-2\left(4-12+9\right)=12-4m\)

\(12-8+24-18-12=-4m\)

\(-2=-4m=>m=\dfrac{1}{2}\)

vậy để pt có nghiệm x=3 thì m=\(\dfrac{1}{2}\)

\(3x^2+4x-4=0\)

\(3x^2+6x-2x-4=0\)

\(3x\left(x+2\right)-2\left(x+2\right)=0\)

\(\left(x+2\right)\left(3x-2\right)=0\)

\(=>x+2=0=>x=-2\)

hoặc \(3x-2=0=>x=\dfrac{2}{3}\)