2x^+3x-1在实数范围内因式分解怎么做

原式=2(x²+3x/2-1/2)
=2(x²+3x/2+9/16-9/16-1/2)
=2[(x+3/4)²-17/16]
=2[(x+3/4)²-(√17/4)²]
=2(x+3/4+√17/4)(x+3/4-√17/4)

2x^+3x-1
=2(x^+3x/2)-1
=2(x^+3x/2+9/16)-1-18/16
=2(x+3/4)^2-34/16
=2[(x+3/4)^2-17/16]
=2(x+3/4-√17/4)(x+3/4+√17/4)

题有问题