1.2[x-3]^2=[2x-5][2x+1]2.2y^2+8y+8=03.-3/2x^2+x+2=0

问题：1.2[x-3]^2=[2x-5][2x+1]2.2y^2+8y+8=03.-3/2x^2+x+2=0

答案：↓↓↓

刘培桐的回答：

网友采纳　　1.2[x-3]^2=[2x-5][2x+1]2x^2-12x+18=4x^2+2x-10x-52x^2+4x-23=0x^2+2x=23/2[x+1]^2=25/2x+1=+/-5/根2x1=-1+5/根2x2=-1-5/根22.2y^2+8y+8=0y^2+4y+4=0[y+2]^2=0y=-23.-3/2x^2+x+2=0x^2-2/3x-4/3=0[x-1/3]^2=4/3...